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Polynomials

2.1  Introduction

  If a is any real number, then a^1=a , a^2=aa ,  a^3=aaa and, in general, if n is any positive integer, the symbol a^n is defined by the equation

   a^n=aa...a ( n factors)

  In the symbol a^n,a is called Hie base and n is called the exponent. The term a^2 is read “ a square,” a^3 is read “ a cubed,” a^4 is read " a to the fourth power," and in general a^n is read “ a to the n th power.”   A variable is a letter that takes on dmerent values from a given collection of real numbers during a given discussion. A constant is a symbol or letter that stands for just one particular real number during the discussion, even if we do not specify which real number it stands for. It is an agreed custom to use the first letters of the alphabet, such as a, b, c, d, ... , for constants and the latter letters of the alphabet. such as x, y, z, u, v, ... , for variables. Since there are only a finite number of letters in the alphabet we are sometimes forced to use subscripts on a single letter to distinguish between different constants. For example, a_0 is read “ a -sub-nought" or “ a -sub-zero," a , is read " a - sub-one,” and, in general, for n a positive integer a_n , is read “ a -sub- n .” We certainly must not confuse subscripts with exponents.

  By a monomial in the variables x, y, ... , z , we mean an expression of the form

   ax^(n)y^(m)...z^(v)

  where n, m, ..., p are positive integers. For example, 7x^3y^2z^5 is a monomial in the variables x, y , and z . Constants are also referred to as monomials. A polynomial in the variables x, y,..., z is any sum of monomials in x, y,..., z . In particular, a binomial is the sum of two monomials and a trinomial is the sum of three monomials. A monomial appearing in a polynomial is referred to as a term of the polynomial.

  By the degree of a monomial we shall mean the sum of the exponents of the variables, or if the monomial is a nonzero constant its degree is understood to be 0 . Thus, the monomial 5 is of degree zero, 3x  is of degree one, while 7x^3y^2z^5 is of degree ten. No degree is assigned to the monomial 0 . Furthermore, while the term 7x^3y^2z^5 is of degree ten, it is also of degree three in x , two in y , and live in z . By the degree of a polynomial, we shall mean the degree of the monomial of highest degree appearing in the polynomial. Polynomials of degree one, two, or three often are called linear, quadratic, or cubic polynomials respectively.

Example 1. Find the degree, the degree in x , and the degree in y of the polynomial 7x^2y^3-4xy^2-x^3y+9y^4 . The terms of the polynomial are the monomials 7x^2y^3,-4xy^2-x^3y , and  9y^4 .

  Consequently the degree of the polynomial in x is 3 , the degree in y is 4 , and its degree is 5 , as indicated in the table above.  

  Any collection of factors in a given monomial is called the coefficient of the remaining factors in the monomial. Thus in the monomial 3xy, 3 is the coefficient of xy , while 3y is the coefficient of x . The numerical factor of a monomial is referred to as the numerical coefficient or simply the coeificienl of the monomial. For example, 3 is the coefficient of 3x^2y, 1 is the coefiicient of x^2y^3 , while -1 is the coefficient of -x

  The leading coefficient of a. polynomial in a single variable is the coefficient of the term of highest degree. The constant term is the term with no variable factor. If we write the polynomial in descending powers of x , as

   a_(n)x^n+a_(n-1)+x^(n-1)+...+a_1x+a_0

  then the polynomial is of degree n , has leading coefficient a_n , and constant term a_0 .

Example 2.  What is the degree, leading coefficient, and constant term of each of the following?

  (a)  3x^2-2x+1

  (b)  7x-4x^3+3

  (c)  2x-4x^2-x^5

  (d)  5

  The value of a polynomial in the variable x at the real number x = a is the real number obtained by replacing each occurrence of x in the polynomial by a. For example, the value of 2x^2- x + 7 at x = -3 is

   2(-3)^2-(-3)+7=2*9+3+7

    = 18+3+7=28

  The value of a polynomial in two or more variables is obtained in a similar way. We obtain the value of xy-x^2+ y^3 at x = 1 ,  y = -2 by replacing each x with 1 and each y with -2 :

   (1)(-2)-(1)^2+(-2)^3=-2-1-8=-11

  Thus the value of  xy-x^2+y^3 at  x=1 ,  y=-2 is  -11 .

2.2  Addition and Subtraction of Polynomials

  Since polynomials are expressions in one or more variables over the real numbers, the laws that we discussed in Chapter 1 may be used to develop techniques for adding, subtracting, multiplying, and dividing them.

  The addition of two expressions, such as 2a and 4a , may be accomplished by a direct application of the distributive law. Thus,

   2a+4a=(2+4)a=6a

  Similarly,

   7c-11c=(7-11)c=-4c

  An expression of the form 2x + 3y cannot be put in any simpler form since in general x and y will denote two different quantities. We can, however, add two or more expressions of this kind in the following way.

Example 1.   Add  2a+3b-4c and  6a-5b+2c .

  Using the associative and commutative laws for addition as well as the distributive law we have

   (2a+3b-4c)+(6a-5b+2c)

  = (2a+6a)+(3b-5b)+(-4c+2c)

  = (2+6)a+(3-5)b+(-4+2)c

  = 8a-2b-2c

  In practice, when several expressions are to be added, the following method is sometimes helpful.

   Add  3a-4ab^3+7c^3 ,  7ab-4a+5c^3 ,  2ab^2-4a+8c^3 , and  -5a+4ab-2ab^2+3c^3 .

  We write the expressions directly underneath one another in such a way that the terms containing the same letters appear in the same column as follows:

addition and subtraction of polynomials

  The bottom line is the final result, which is obtained by adding the respective columns.

  We have already seen (Chapter 1) that subtraction of signed numbers may be accomplished by addition after changing the sign of the number to be subtracted. This is exactly what we do when we subtract polynomials.

   Subtract  3x-2s+t from  4x+s-2t

   (4x+s-2t)-(3x-2s+t)

  = 4x+s-2t-3x+2s-t

  = (4x-3x)+(s+2s)+(-2t-t)

  = x+3s-3t

   From the sum of 2a + 7b - 15c and 60 - 4b + c subtract the sum of a-b+2c and -2a+6b-3c .   Since subtracting the sum of a - b + 2c and - -2a+6b-3c is the same as subtracting each expression separately, we simply change all the signs of the last two expressions and add.

addition and subtraction of polynomials - 1

  We handle grouping with parentheses the same way that we handled it with signed numbers in Section 1.2.

   Simplify  8b-{7a-[(a-3b)-(2a+5b)]} .

   8b-{7a-[(a-3b)-(2a+5b)]}

  = 8b-{7a-[a-3b-2a-5b]}

  = 8b-{7a+[-a-8b]}

  = 8b-{7a+a+8b}

  = 8b-{8a+8b}

  = -8a

Let’s see how our Polynomials solver simplifies this and similar problems. Click on "Solve Similar" button to see more examples.

   Rewrite the expression x-2y+ z-5 with the last three terms enclosed in parentheses preceded by a minus sign.

   x-2y+z-5=x-(2y-z+5)

2.3  Multiplication of Polynomials

   When multiplying monomials in which the variable x appears, we obtain products of the form  x^(m)x^(n) . The total number of factors of x in this product is m + n , so that we have the following law of exponents:

   x^(m)x^(n)=x^(m+n)

  For example,

   x^(2)x^(7)=x^(2+7)=x^9

  In order to multiply two or more algebraic expressions together we must make use of the above law as well as the laws of real numbers from Chapter 1.

  Simplify  (-3x^2y^3z)(2x^4yz^3) .

   (-3x^2y^3z)(2x^4yz^3)

  = (-3*2)(x^2x^4)(y^3y)(z*z^3)

  = -6x^6y^4z^4

  In particular we make use of the distributive laws when we multiply two multinomials, as is illustrated in the following examples.

   Multiply  (x^3+2x^2-x+2) by  (2x)

   (x^3+2x^2-x+2)(2x)

  = x^3(2x)+2x^3(2x)-x(2x)+2(2x)

  = 2x^4+4x^3-2x^2+4x

   Multiply  3x^2+x-5 by  x+2 .

   (3x^2+x-5)(x+2)

  = 3x^2(x+2)+x(x+2)-5(x+2)

  = 3x^2(x)+3x^2(2)+x(x)+x(2)-5(x)-5(2)

  = 3x^3+6x^2+x^2+2x-5x-10

  = 3x^3+7x^2-3x-10

  From Example 3 we see that the terms in the product of one polynomial by another are obtained by multiplying each term in the first factor by each term in the second. With a little experience we will be able to skip the steps that use the distributive laws, simply writing down all the products. This is illustrated in the next two examples.

Let’s see how our polynomial solver Multiplies this and similar problems. Click on "Solve Similar" button to see more examples.

   Multiply  2x+3 by  x-5

   (2x+3)(x-5)

  = 2x(x)+2x(-5)+2(x)+3(-5)

  = 2x^3-10x+3x-15

  = 2x^3-7x-15

   Multiply  2x+y by  x+2y-3 .

   (2x+y)(x+2y-3)

  = 2x^2+4xy-6x+xy+2y^2-3y

  = 2x^2+5xy+2y^2-6x-3y

2.4  Division of Polynomials

  In order to divide one polynomial by another, we first must be able to divide one monomial by another. When dividing one monomial in at by another we must consider expressions of the form x^(m)÷x^(n) where m > n . This quotient is x^(m-n) since x^(m-n)x^n=x^m . Also if m = n , then x^m÷x^n = 1 . This can be summarized as another law of exponents:

Multiplication of polynomials

   x^7÷x^2=x^(7-2)=x^5

  Since

   x^5x^2=x^(5+2)=x^7

   Divide  8x^4y^2z^3 by  2x^2yz .

  We use the notation of long division, which is what we will find convenient for dividing one polynomial by another.

Division of polynomials

  Thus,

   8x^4y^2z^3÷2x^2yz=4x^2yz^3

  The next example recalls the method of long division for integers.

   Divide  492 by  8 .

Division of polynomials - 1

  This gives us the equality

   492=(61)(8)+4

  Recall that 492 is called the dividend, 8 the divisor, 61 the quotient, and 4 the remainder. Note the division process terminates when the remainder is less than the divisor.

  The division of one polynomial by another is carried out in a similar manner. Here the division terminates when the degree of the remainder is less than the degree of the divisor, or when the remainder is zero.

   Divide  -3x+2x^2+5 by  x-1 .

Step 1.   Arrange both polynomials in descending powers of x , and write as follows.

Division of polynomials - 2

Step 2.   Divide  x into  2x^2 .

Division of polynomials - 3

Step 3.   Multiply  x-1 by  2x and subtract from  2x^2-3x+5 .

Division of polynomials - 4

  Since the degree of the remainder, -x+5 , is not less than the degree of the divisor, x-1 , we repeat the process.

Step 4.   Divide  x into  -x .

Division of polynomials - 5

Step 5.   Multiply  x-1 by  -1 and subtract from  -x+5 .

Division of polynomials - 6

  The remainder, 4 , has degree 0 , and Hie divisor, x- 1 , has degree 1 Therefore, the division terminates.

Step 6.   The result of this division is presented by the equation

    (dividend) = (quotient)(divisor) + (remainder)

  which in this example is

   2x^2-3x+5=(2x-1)(x-1)+4

   Divide  3x^3-2x+5 by  x^2+x+1 .

Division of polynomials - 7

  Since the degree of the remainder, -2x+8 , is less than the degree of the divisor, x^2+x + 1 , the process terminates. The result is

   3x^2-2x+5 = (3x-3)(x^2+x+1)-2x+8 .

   Divide  3x^3+2x^2-1 by  2x+7 .

Division of polynomials - 8

   3x^3+2x^2-1 = (3/2x^2-17/4x+119/8)(2x+7)-841/8

   Divide  x^3+a^3 by  x-a

  To perform this division we treat these polynomials as polynomials in the single variable x .

Division of polynomials - 9

   x^3+a^3 = (x^2+ax+a^2)(x-a)+2a^3

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Solving Polynomials

"Solving" means finding the "roots" ...

... a "root" (or "zero") is where the function is equal to zero :

In between the roots the function is either entirely above, or entirely below, the x-axis

Example: −2 and 2 are the roots of the function x 2 − 4

Let's check:

How do we solve polynomials? That depends on the Degree !

The first step in solving a polynomial is to find its degree.

The Degree of a Polynomial with one variable is ...

... the largest exponent of that variable.

When we know the degree we can also give the polynomial a name:

How To Solve

So now we know the degree, how to solve?

So what do we do with ones we can't solve? Try to solve them a piece at a time!

If we find one root, we can then reduce the polynomial by one degree (example later) and this may be enough to solve the whole polynomial.

Here are some main ways to find roots.

1. Basic Algebra

We may be able to solve using basic algebra:

Example: 2x+1

2x+1 is a linear polynomial:

The graph of y = 2x+1 is a straight line

It is linear so there is one root.

Use Algebra to solve:

A "root" is when y is zero: 2x+1 = 0

Subtract 1 from both sides: 2x = −1

Divide both sides by 2: x = −1/2

And that is the solution:

(You can also see this on the graph)

We can also solve Quadratic Polynomials using basic algebra (read that page for an explanation).

2. By experience, or simply guesswork.

It is always a good idea to see if we can do simple factoring:

Example: x 3 +2x 2 −x

This is cubic ... but wait ... we can factor out "x":

x 3 +2x 2 −x = x(x 2 +2x−1)

Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.

Example: x 3 −8

Again this is cubic ... but it is also the " difference of two cubes ":

x 3 −8 = x 3 −2 3

And so we can turn it into this:

x 3 −8 = (x−2)(x 2 +2x+4)

There is a root at x=2, because:

(2−2)(2 2 +2×2+4) = (0) (2 2 +2×2+4)

And we can then solve the quadratic x 2 +2x+4 and we are done

3. Graphically.

Graph the polynomial and see where it crosses the x-axis.

Graphing is a good way to find approximate answers, and we may also get lucky and discover an exact answer.

Caution: before you jump in and graph it, you should really know How Polynomials Behave , so you find all the possible answers!

This is useful to know: When a polynomial is factored like this:

f(x) = (x−a)(x−b)(x−c)...

Then a, b, c, etc are the roots !

So Linear Factors and Roots are related, know one and we can find the other.

(Read The Factor Theorem for more details.)

Example: f(x) = (x 3 +2x 2 )(x−3)

We see "(x−3)", and that means that 3 is a root (or "zero") of the function.

Well, let us put "3" in place of x:

f(x) = (3 3 +2·3 2 )(3−3)

f(3) = (3 3 +2·3 2 )( 0 )

Yes! f(3)=0, so 3 is a root.

How to Check

Found a root? Check it!

Simply put the root in place of "x": the polynomial should be equal to zero.

Example: 2x 3 −x 2 −7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2 . We can check easily, just put "2" in place of "x":

f(2) = 2(2) 3 −(2) 2 −7(2)+2 = 16−4−14+2 = 0

Yes! f(2)=0 , so we have found a root!

How about where it crosses near −1.8 :

f(−1.8) = 2(−1.8) 3 −(−1.8) 2 −7(−1.8)+2 = −11.664−3.24+12.6+2 = −0.304

No, it isn't equal to zero, so −1.8 will not be a root (but it may be close!)

But we did discover one root, and we can use that to simplify the polynomial, like this

Example (continued): 2x 3 −x 2 −7x+2

So, f(2)=0 is a root ... that means we also know a factor:

(x−2) must be a factor of 2x 3 −x 2 −7x+2

Next, divide 2x 3 −x 2 −7x+2 by (x−2) using Polynomial Long Division to find:

2x 3 −x 2 −7x+2 = (x−2)(2x 2 +3x−1)

So now we can solve 2x 2 +3x−1 as a Quadratic Equation and we will know all the roots.

That last example showed how useful it is to find just one root. Remember:

If we find one root, we can then reduce the polynomial by one degree and this may be enough to solve the whole polynomial.

How Far Left or Right

When trying to find roots, how far left and right of zero should we go?

There is a way to tell, and there are a few calculations to do, but it is all simple arithmetic. Read Bounds on Zeros for all the details.

Have We Got All The Roots?

There is an easy way to know how many roots there are. The Fundamental Theorem of Algebra says:

A polynomial of degree n ... ... has n roots (zeros) but we may need to use complex numbers

So: number of roots = the degree of polynomial .

Example: 2x 3 + 3x − 6

The degree is 3 (because the largest exponent is 3), and so:

There are 3 roots.

But Some Roots May Be Complex

Yes, indeed, some roots may be complex numbers (ie have an imaginary part), and so will not show up as a simple "crossing of the x-axis" on a graph.

But there is an interesting fact:

Complex Roots always come in pairs !

So we either get no complex roots, or 2 complex roots, or 4 , etc... Never an odd number.

Which means we automatically know this:

Positive or Negative Roots?

There is also a special way to tell how many of the roots are negative or positive called the Rule of Signs that you may like to read about.

Multiplicity of a Root

Sometimes a factor appears more than once. We call that Multiplicity :

Example: f(x) = (x−5) 3 (x+7)(x−1) 2

This could be written out in a more lengthy way like this:

f(x) = (x−5)(x−5)(x−5)(x+7)(x−1)(x−1)

(x−5) is used 3 times, so the root "5" has a multiplicity of 3 , likewise (x+7) appears once and (x−1) appears twice. So:

Q: Why is this useful? A: It makes the graph behave in a special way!

When we see a factor like (x-r) n , "n" is the multiplicity, and

We can see it on this graph:

Example: f(x) = (x−2) 2 (x−4) 3

(x−2) has even multiplicity , so it just touches the axis at x=2

(x−4) has odd multiplicity , so it crosses the axis at x=4

solve math problems polynomials

4.4 Solve Polynomial Equations by Factoring

Learning objectives.

Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials:

Determine the number of terms in the polynomial.

Factor binomials (2 terms) using the following special products:

Note : If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization. In addition, not all polynomials with integer coefficients factor. When this is the case, we say that the polynomial is prime.

If an expression has a GCF, then factor this out first. Doing so is often overlooked and typically results in factors that are easier to work with. Furthermore, look for the resulting factors to factor further; many factoring problems require more than one step. A polynomial is completely factored when none of the factors can be factored further.

Factor: 54 x 4 − 36 x 3 − 24 x 2 + 16 x .

This four-term polynomial has a GCF of 2 x . Factor this out first.

54 x 4 − 36 x 3 − 24 x 2 + 16 x = 2 x ( 27 x 3 − 18 x 2 − 12 x + 8 )

Now factor the resulting four-term polynomial by grouping and look for resulting factors to factor further.

solve math problems polynomials

Answer: 2 x ( 3 x − 2 ) 2 ( 3 x + 2 ) . The check is left to the reader.

Factor: x 4 − 3 x 2 − 4 .

This trinomial does not have a GCF.

x 4 − 3 x 2 − 4 = ( x 2             ) ( x 2             ) = ( x 2 + 1 ) ( x 2 − 4 )          D i f f e r e n c e   o f   s q u a r e s = ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

The factor ( x 2 + 1 ) is prime and the trinomial is completely factored.

Answer: ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

Factor: x 6 + 6 x 3 − 16 .

Begin by factoring x 6 = x 3 ⋅ x 3 and look for the factors of 16 that add to 6.

The factor ( x 3 − 2 ) cannot be factored any further using integers and the factorization is complete.

Answer: ( x 3 − 2 ) ( x + 2 ) ( x 2 + 2 x + 4 )

Try this! Factor: 9 x 4 + 17 x 2 − 2

Answer: ( 3 x + 1 ) ( 3 x − 1 ) ( x 2 + 2 )

Solving Polynomial Equations by Factoring

In this section, we will review a technique that can be used to solve certain polynomial equations. We begin with the zero-product property A product is equal to zero if and only if at least one of the factors is zero. :

a ⋅ b = 0    if and only if    a = 0  or  b = 0

The zero-product property is true for any number of factors that make up an equation. In other words, if any product is equal to zero, then at least one of the variable factors must be equal to zero. If an expression is equal to zero and can be factored into linear factors, then we will be able to set each factor equal to zero and solve for each equation.

Solve: 2 x ( x − 4 ) ( 5 x + 3 ) = 0 .

Set each variable factor equal to zero and solve.

2 x = 0   or   x − 4 = 0   or   5 x + 3 = 0 2 x 2 = 0 2 x = 4 5 x 5 = − 3 5 x = 0 x = − 3 5

To check that these are solutions we can substitute back into the original equation to see if we obtain a true statement. Note that each solution produces a zero factor. This is left to the reader.

Answer: The solutions are 0, 4, and − 3 5 .

Of course, most equations will not be given in factored form.

Solve: 4 x 3 − x 2 − 100 x + 25 = 0 .

Begin by factoring the left side completely.

4 x 3 − x 2 − 100 x + 25 = 0 F a c t o r   b y   g r o u p i n g . x 2 ( 4 x − 1 ) − 25 ( 4 x − 1 ) = 0 ( 4 x − 1 ) ( x 2 − 25 ) = 0 F a c t o r   a s   a   d i f f e r e n c e   o f   s q u a r e s . ( 4 x − 1 ) ( x + 5 ) ( x − 5 ) = 0

Set each factor equal to zero and solve.

4 x − 1 = 0 or x + 5 = 0 or x − 5 = 0 4 x = 1 x = − 5 x = 5 x = 1 4

Answer: The solutions are 1 4 , −5, and 5.

Using the zero-product property after factoring an equation that is equal to zero is the key to this technique. However, the equation may not be given equal to zero, and so there may be some preliminary steps before factoring. The steps required to solve by factoring The process of solving an equation that is equal to zero by factoring it and then setting each variable factor equal to zero. are outlined in the following example.

Solve: 15 x 2 + 3 x − 8 = 5 x − 7 .

Step 1: Express the equation in standard form, equal to zero. In this example, subtract 5 x from and add 7 to both sides.

15 x 2 + 3 x − 8 = 5 x − 7 15 x 2 − 2 x − 1 = 0

Step 2: Factor the expression.

( 3 x − 1 ) ( 5 x + 1 ) = 0

Step 3: Apply the zero-product property and set each variable factor equal to zero.

3 x − 1 = 0         or         5 x + 1 = 0

Step 4: Solve the resulting linear equations.

3 x − 1 = 0 or 5 x + 1 = 0 3 x = 1 5 x = − 1 x = 1 3 x = − 1 5

Answer: The solutions are 1 3 and − 1 5 . The check is optional.

Solve: ( 3 x + 2 ) ( x + 1 ) = 4 .

This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to 4. However, this would lead to incorrect results. We must rewrite the equation equal to zero, so that we can apply the zero-product property.

( 3 x + 2 ) ( x + 1 ) = 4 3 x 2 + 3 x + 2 x + 2 = 4 3 x 2 + 5 x + 2 = 4 3 x 2 + 5 x − 2 = 0

Once it is in standard form, we can factor and then set each factor equal to zero.

                ( 3 x − 1 ) ( x + 2 ) = 0 3 x − 1 = 0               or           x + 2 = 0               3 x = 1                                                 x = − 2                     x = 1 3

Answer: The solutions are 1 3 and −2.

Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0

A root A value in the domain of a function that results in zero. of a function is a value in the domain that results in zero. In other words, the roots occur when the function is equal to zero, f ( x ) = 0 .

Find the roots: f ( x ) = ( x + 2 ) 2 − 4 .

To find roots we set the function equal to zero and solve.

f ( x ) = 0 ( x + 2 ) 2 − 4 = 0 x 2 + 4 x + 4 − 4 = 0 x 2 + 4 x = 0 x ( x + 4 ) = 0

Next, set each factor equal to zero and solve.

x = 0   or     x + 4 = 0 x = − 4

We can show that these x -values are roots by evaluating.

f ( 0 ) = ( 0 + 2 ) 2 − 4         f ( − 4 ) = ( − 4 + 2 ) 2 − 4 = 4 − 4   = ( − 2 ) 2 − 4 = 0         ✓ = 4 − 4 = 0         ✓

Answer: The roots are 0 and −4.

If we graph the function in the previous example we will see that the roots correspond to the x -intercepts of the function. Here the function f is a basic parabola shifted 2 units to the left and 4 units down.

solve math problems polynomials

Find the roots: f ( x ) = x 4 − 5 x 2 + 4 .

f ( x ) = 0 x 4 − 5 x 2 + 4 = 0 ( x 2 − 1 ) ( x 2 − 4 ) = 0 ( x + 1 ) ( x − 1 ) ( x + 2 ) ( x − 2 ) = 0

x + 1 = 0         or         x − 1 = 0         or         x + 2 = 0         or         x − 2 = 0 x = − 1 x = 1 x = − 2 x = 2

Answer: The roots are −1, 1, −2, and 2.

Graphing the previous function is not within the scope of this course. However, the graph is provided below:

solve math problems polynomials

Notice that the degree of the polynomial is 4 and we obtained four roots. In general, for any polynomial function with one variable of degree n , the fundamental theorem of algebra Guarantees that there will be as many (or fewer) roots to a polynomial function with one variable as its degree. guarantees n real roots or fewer. We have seen that many polynomials do not factor. This does not imply that functions involving these unfactorable polynomials do not have real roots. In fact, many polynomial functions that do not factor do have real solutions. We will learn how to find these types of roots as we continue in our study of algebra.

Find the roots: f ( x ) = − x 2 + 10 x − 25 .

f ( x ) = 0 − x 2 + 10 x − 25 = 0 − ( x 2 − 10 x + 25 ) = 0 − ( x − 5 ) ( x − 5 ) = 0

Next, set each variable factor equal to zero and solve.

x − 5 = 0 or x − 5 = 0 = 5 x = 5

A solution that is repeated twice is called a double root A root that is repeated twice. . In this case, there is only one solution.

Answer: The root is 5.

The previous example shows that a function of degree 2 can have one root. From the factoring step, we see that the function can be written

f ( x ) = − ( x − 5 ) 2

In this form, we can see a reflection about the x -axis and a shift to the right 5 units. The vertex is the x -intercept, illustrating the fact that there is only one root.

solve math problems polynomials

Try this! Find the roots of f ( x ) = x 3 + 3 x 2 − x − 3 .

Answer: ±1, −3

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x , where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 40 feet.

We are asked to find the speed x where the safe stopping distance d ( x ) = 40 feet.

d ( x ) = 40 1 20 x 2 + x = 40

To solve for x , rewrite the resulting equation in standard form. In this case, we will first multiply both sides by 20 to clear the fraction.

20 ( 1 20 x 2 + x ) = 20 ( 40 ) x 2 + 20 x = 800 x 2 + 20 x − 800 = 0

Next factor and then set each factor equal to zero.

x 2 + 20 x − 800 = 0 ( x + 40 ) ( x − 20 ) = 0 x + 40 = 0 o r x − 20 = 0 x = − 40 x = 20

The negative answer does not make sense in the context of this problem. Consider x = 20 miles per hour to be the only solution.

Answer: 20 miles per hour

Finding Equations with Given Solutions

We can use the zero-product property to find equations, given the solutions. To do this, the steps for solving by factoring are performed in reverse.

Find a quadratic equation with integer coefficients, given solutions − 3 2 and 1 3 .

Given the solutions, we can determine two linear factors. To avoid fractional coefficients, we first clear the fractions by multiplying both sides by the denominator.

x = − 3 2 or x = 1 3 2 x = − 3 3 x = 1 2 x + 3 = 0 3 x − 1 = 0

The product of these linear factors is equal to zero when x = − 3 2 or x = 1 3 .

( 2 x + 3 ) ( 3 x − 1 ) = 0

Multiply the binomials and present the equation in standard form.

6 x 2 − 2 x + 9 x − 3 = 0 6 x 2 + 7 x − 3 = 0

We may check our equation by substituting the given answers to see if we obtain a true statement. Also, the equation found above is not unique and so the check becomes essential when our equation looks different from someone else’s. This is left as an exercise.

Answer: 6 x 2 + 7 x − 3 = 0

Find a polynomial function with real roots 1, −2, and 2.

Given solutions to f ( x ) = 0 we can find linear factors.

x = 1 or x = − 2 or x = 2 x − 1 = 0 x + 2 = 0 x − 2 = 0

Apply the zero-product property and multiply.

( x − 1 ) ( x + 2 ) ( x − 2 ) = 0 ( x − 1 ) ( x 2 − 4 ) = 0 x 3 − 4 x − x 2 + 4 = 0 x 3 − x 2 − 4 x + 4 = 0

Answer: f ( x ) = x 3 − x 2 − 4 x + 4

Try this! Find a polynomial equation with integer coefficients, given solutions 1 2 and − 3 4 .

Answer: 8 x 2 + 2 x − 3 = 0

Key Takeaways

Topic Exercises

Part a: general factoring.

Factor completely.

50 x 2 − 18

12 x 3 − 3 x

10 x 3 + 65 x 2 − 35 x

15 x 4 + 7 x 3 − 4 x 2

6 a 4 b − 15 a 3 b 2 − 9 a 2 b 3

8 a 3 b − 44 a 2 b 2 + 20 a b 3

36 x 4 − 72 x 3 − 4 x 2 + 8 x

20 x 4 + 60 x 3 − 5 x 2 − 15 x

3 x 5 + 2 x 4 − 12 x 3 − 8 x 2

10 x 5 − 4 x 4 − 90 x 3 + 36 x 2

x 4 − 23 x 2 − 50

2 x 4 − 31 x 2 − 16

− 2 x 5 − 6 x 3 + 8 x

− 36 x 5 + 69 x 3 + 27 x

54 x 5 − 78 x 3 + 24 x

4 x 6 − 65 x 4 + 16 x 2

x 6 − 7 x 3 − 8

x 6 − 25 x 3 − 54

3 x 6 + 4 x 3 + 1

27 x 6 − 28 x 3 + 1

Part B: Solving Polynomial Equations by Factoring

( 6 x − 5 ) ( x + 7 ) = 0

( x + 9 ) ( 3 x − 8 ) = 0

5 x ( 2 x − 5 ) ( 3 x + 1 ) = 0

4 x ( 5 x − 1 ) ( 2 x + 3 ) = 0

( x − 1 ) ( 2 x + 1 ) ( 3 x − 5 ) = 0

( x + 6 ) ( 5 x − 2 ) ( 2 x + 9 ) = 0

( x + 4 ) ( x − 2 ) = 16

( x + 1 ) ( x − 7 ) = 9

( 6 x + 1 ) ( x + 1 ) = 6

( 2 x − 1 ) ( x − 4 ) = 39

x 2 − 15 x + 50 = 0

x 2 + 10 x − 24 = 0

3 x 2 + 2 x − 5 = 0

2 x 2 + 9 x + 7 = 0

1 10 x 2 − 7 15 x − 1 6 = 0

1 4 − 4 9 x 2 = 0

6 x 2 − 5 x − 2 = 30 x + 4

6 x 2 − 9 x + 15 = 20 x − 13

5 x 2 − 23 x + 12 = 4 ( 5 x − 3 )

4 x 2 + 5 x − 5 = 15 ( 3 − 2 x )

( x + 6 ) ( x − 10 ) = 4 ( x − 18 )

( x + 4 ) ( x − 6 ) = 2 ( x + 4 )

4 x 3 − 14 x 2 − 30 x = 0

9 x 3 + 48 x 2 − 36 x = 0

1 3 x 3 − 3 4 x = 0

1 2 x 3 − 1 50 x = 0

− 10 x 3 − 28 x 2 + 48 x = 0

− 2 x 3 + 15 x 2 + 50 x = 0

2 x 3 − x 2 − 72 x + 36 = 0

4 x 3 − 32 x 2 − 9 x + 72 = 0

45 x 3 − 9 x 2 − 5 x + 1 = 0

x 3 − 3 x 2 − x + 3 = 0

x 4 − 5 x 2 + 4 = 0

4 x 4 − 37 x 2 + 9 = 0

Find the roots of the given functions.

f ( x ) = x 2 + 10 x − 24

f ( x ) = x 2 − 14 x + 48

f ( x ) = − 2 x 2 + 7 x + 4

f ( x ) = − 3 x 2 + 14 x + 5

f ( x ) = 16 x 2 − 40 x + 25

f ( x ) = 9 x 2 − 12 x + 4

g ( x ) = 8 x 2 + 3 x

g ( x ) = 5 x 2 − 30 x

p ( x ) = 64 x 2 − 1

q ( x ) = 4 x 2 − 121

f ( x ) = 1 5 x 3 − 1 x 2 − 1 20 x + 1 4

f ( x ) = 1 3 x 3 + 1 2 x 2 − 4 3 x − 2

g ( x ) = x 4 − 13 x 2 + 36

g ( x ) = 4 x 4 − 13 x 2 + 9

f ( x ) = ( x + 5 ) 2 − 1

g ( x ) = − ( x + 5 ) 2 + 9

f ( x ) = − ( 3 x − 5 ) 2

g ( x ) = − ( x + 2 ) 2 + 4

Given the graph of a function, determine the real roots.

solve math problems polynomials

The sides of a square measure x − 2 units. If the area is 36 square units, then find x .

The sides of a right triangle have lengths that are consecutive even integers. Find the lengths of each side. (Hint: Apply the Pythagorean theorem)

The profit in dollars generated by producing and selling n bicycles per week is given by the formula P ( n ) = − 5 n 2 + 400 n − 6000 . How many bicycles must be produced and sold to break even?

The height in feet of an object dropped from the top of a 64-foot building is given by h ( t ) = − 16 t 2 + 64 where t represents the time in seconds after it is dropped. How long will it take to hit the ground?

A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box of height 2 inches is given.

solve math problems polynomials

What is the length of each side of the cardboard sheet if the volume of the box is to be 98 cubic inches?

The height of a triangle is 4 centimeters less than twice the length of its base. If the total area of the triangle is 48 square centimeters, then find the lengths of the base and height.

A uniform border is to be placed around an 8 × 10 inch picture.

solve math problems polynomials

If the total area including the border must be 168 square inches, then how wide should the border be?

The area of a picture frame including a 3-inch wide border is 120 square inches.

solve math problems polynomials

If the width of the inner area is 2 inches less than its length, then find the dimensions of the inner area.

Assuming dry road conditions and average reaction times, the safe stopping distance in feet is given by d ( x ) = 1 20 x 2 + x where x represents the speed of the car in miles per hour. Determine the safe speed of the car if you expect to stop in 75 feet.

A manufacturing company has determined that the daily revenue in thousands of dollars is given by the formula R ( n ) = 12 n − 0.6 n 2 where n represents the number of palettes of product sold ( 0 ≤ n < 20 ) . Determine the number of palettes sold in a day if the revenue was 45 thousand dollars.

Part C: Finding Equations with Given Solutions

Find a polynomial equation with the given solutions.

Find a function with the given roots.

2 5 , − 1 3

5 double root

−3 double root

Recall that if | X | = p , then X = − p or X = p . Use this to solve the following absolute value equations.

| x 2 − 8 | = 8

| 2 x 2 − 9 | = 9

| x 2 − 2 x − 1 | = 2

| x 2 − 8 x + 14 | = 2

| 2 x 2 − 4 x − 7 | = 9

| x 2 − 3 x − 9 | = 9

Part D: Discussion Board

Explain to a beginning algebra student the difference between an equation and an expression.

What is the difference between a root and an x -intercept? Explain.

Create a function with three real roots of your choosing. Graph it with a graphing utility and verify your results. Share your function on the discussion board.

Research and discuss the fundamental theorem of algebra.

2 ( 5 x + 3 ) ( 5 x − 3 )

5 x ( x + 7 ) ( 2 x − 1 )

3 a 2 b ( 2 a + b ) ( a − 3 b )

4 x ( x − 2 ) ( 3 x + 1 ) ( 3 x − 1 )

x 2 ( 3 x + 2 ) ( x + 2 ) ( x − 2 )

( x 2 + 2 ) ( x + 5 ) ( x − 5 )

6 x ( x + 1 ) ( x − 1 ) ( 3 x + 2 ) ( 3 x − 2 )

( x + 1 ) ( x 2 − x + 1 ) ( x − 2 ) ( x 2 + 2 x + 4 )

( 3 x 3 + 1 ) ( x + 1 ) ( x 2 − x + 1 )

0, 5 2 , − 1 3

− 1 2 , 1, 5 3

− 5 3 , 1 2

0, − 3 2 , 5

± 1 3 , 1 5

−3, −1, 0, 2

20 or 60 bicycles

30 miles per hour

x 2 − 2 x − 15 = 0

3 x 2 − 7 x + 2 = 0

x 2 + 4 x = 0

x 2 − 49 = 0

x 3 − x 2 − 9 x + 9 = 0

f ( x ) = 6 x 2 − 7 x + 2

f ( x ) = 16 x 2 − 9

f ( x ) = x 2 − 10 x + 25

f ( x ) = x 3 − 2 x 2 − 3 x

Answer may vary

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How to Solve Polynomials

Last Updated: October 18, 2022 References

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 12 references cited in this article, which can be found at the bottom of the page. This article has been viewed 297,269 times.

A polynomial is an expression made up of adding and subtracting terms. A terms can consist of constants, coefficients, and variables. When solving polynomials, you usually trying to figure out for which x-values y=0. Lower-degree polynomials will have zero, one or two real solutions, depending on whether they are linear polynomials or quadratic polynomials. These types of polynomials can be easily solved using basic algebra and factoring methods. For help solving polynomials of a higher degree, read Solve Higher Degree Polynomials .

Solving a Linear Polynomial

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About This Article

David Jia

To solve a linear polynomial, set the equation to equal zero, then isolate and solve for the variable. A linear polynomial will have only one answer. If you need to solve a quadratic polynomial, write the equation in order of the highest degree to the lowest, then set the equation to equal zero. Rewrite the expression as a 4-term expression and factor the equation by grouping. Rewrite the polynomial as 2 binomials and solve each one. If you want to learn how to simplify and solve your terms in a polynomial equation, keep reading the article! Did this summary help you? Yes No

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Solving Polynomials

Solving Factoring Examples

The general technique for solving bigger-than-quadratic polynomials is pretty straightforward, but the process can be time-consuming.

Note: The terminology for this topic is often used carelessly. Technically, one "solves" an equation, such as "(polynomal) equals (zero)"; one "finds the roots" of a function, such as "( y ) equals (polynomial)". On this page, regardless of how the topic is framed, the point will be to find all of the solutions to "(polynomial) equals (zero)", even if the question is stated differently, such as "Find the roots of ( y ) equals (polynomial)".

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The first step in finding the solutions of (that is, the x -intercepts of, plus any complex -valued roots of) a given polynomial function is to apply the Rational Roots Test to the polynomial's leading coefficient and constant term, in order to get a list of values that might possibly be solutions to the related polynomial equation. Your hand-in work is probably expected to contain this list, so write this out neatly.

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You can follow this up with an application of Descartes' Rule of Signs , if you like, to narrow down which possible zeroes might be best to check. On the other hand, if you've got a graphing calculator you can use, it's easy to do a graph. The x - intercepts of the graph are the same as the (real-valued) zeroes of the equation. Seeing where the line looks as though it crosses the x -axis can quickly narrow down your list of possible zeroes that you'll want first to check.

Once you've found an x -value that you want to test, you then use synthetic division to see if you can get a zero remainder. If you do get a zero remainder, then you've not only found a zero of the original polynomial, but you've also reduced your polynomial by one degree, by effectively removing one factor.

Remember that synthetic division is, among other things, a form of polynomial division, so checking if x  =  a is a solution to "(polynomial) equals (zero)" is the same as dividing the linear factor x  −  a out of the related polynomial function "( y ) equals (polynomial)".

This also means that, after a successful division, you've also successfully taken a factor out. You should not then return to the original polynomial for your next computation for finding the other zeroes. You should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.

(This method will be demonstrated in the examples below.)

You should not be surprised to see some complicated solutions to your polynomials (that is, solutions containing square roots or complex numbers, or both); these zeroes will come from applying the Quadratic Formula to (what is usually) the final (quadratic) factor of your polynomial. You should expect that the answers will be messy.

Here's how the process plays out in practice:

Find all the zeroes of: y = 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x + 24

First, I'll apply the Rational Roots Test—

Wait. Actually, the first thing I'll do is apply a trick I've learned. First, I'll check to see if either x  = 1 or x  = −1 is a root.

(These are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because (a) it's amazing how often one of these is a zero, and (b) you can just look at the powers and the numbers to figure out if either works, because of how 1 and −1 simplify.)

When x  = 1 , the polynomial evaluates as:

2 + 3 − 30 − 57 − 2 + 24 = −60

This isn't equal to zero, so x  = 1 isn't a root. But when x  = −1 , I get:

−2 + 3 + 30 − 57 + 2 + 24 = 0

This time, it did equal zero, so now I know that x  = −1 is a root, and I can take "prove" this (in my hand-in work) by using synthetic division:

The last line of this division shows me with the new, smaller polynomial equation I'm working with now:

2 x 4 + x 3 − 31 x 2 − 26 x + 24 = 0

(I'd started with a degree-five polynomial. Since I've effectively divided out the factor x  + 1 , I've reduced the degree of the polynomial by 1 . That's how I know the last line of the division represents a degree-four polynomial.)

I've taken care of checking the two easiest zeroes. Now I'll apply the Rational Roots Test to what's left in order to get a list of potential zeroes to try:

From experience (mostly by having worked extra homework problems), I've learned that most of these exercises have their zeroes somewhere near the middle of the list, rather than at the extremes. This isn't always true, of course, but it's usually better to stay away from the larger numbers, at least when I'm getting started.

So, in this case, I won't start off by trying stuff like x  = −24 or x  = 12 . Instead, I'll start out with smaller values like x  = 2 .

And I can narrow down my options further by "cheating" and looking at the graph:

This is a fourth-degree polynomial, so it has, at most, four x -intercepts, and I can see all four of them on the graph. This means that I won't have any complex-valued zeroes.

It also looks like there may be zeroes near −1.5 and 0.5 . But the clearest solution looks to be at x  = 4 and since whole numbers are easier to work with than fractions, x  = 4 would probably be a good next value to try:

The zero remainder (at the far right of the bottom row) tells me that x  = 4 is indeed a root. And the bottom row of the synthetic division tells me that I'm now left with solving the following:

2 x 3 + 9 x 2 + 5 x − 6 = 0

Looking at the constant term " 6 " in the polynomial above, and with the Rational Roots Test in mind, I can see that the following values:

x = ±24, ±12, ±8, −4

...from my original application of the Rational Roots Test won't work for the current polynomial. Even if I didn't already know this from having checked the graph, I can see that they won't fit with the new polynomial's leading coefficient and constant term. So I can cross these values off of my list now.

(Always check the list of possible zeroes as you go. The Rational Roots Test will sometimes give a very long list of possibilities, and it can be helpful to notice that some of those values can be ignored, especially if you don't have a graphing calculator to "cheat" with.)

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2 x 2 + 6 x − 4 = 0

Dividing through by 2 to get smaller numbers gives me:

x 2 + 3 x − 2 = 0

I can apply the Quadratic Formula to this:

This gives me the remaining two roots of the original polynomial function. (I plugged the exact values into my calculator, to confirm that they match up with what I'd already seen on the graph, so I'd be certain that my answer was correct. I won't hand in these approximations, though.)

My complete answer is:

Asking you to find the zeroes of a polynomial function, y equals (polynomial), means the same thing as asking you to find the solutions to a polynomial equation, (polynomial) equals (zero). The zeroes of a polynomial are the values of x that make the polynomial equal to zero. Either task may be referred to as "solving the polynomial".

So the above problem could have been stated along the lines of:

Find the solutions to 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x + 24 = 0

Find the solutions to 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x = −24

...and the answers would have been the exact same list of x -values.

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Polynomial Word Problems

Solution of exercise solved polynomial word problems, solution of exercise 1, solution of exercise 2, solution of exercise 3, solution of exercise 4, solution of exercise 5, solution to exercise 6, solution to exercise 7, solution to exercise 8, solution to exercise 9.

A polynomial is an expression which consists of two or more than two algebraic expressions.  In a polynomial expression, the same variable has different powers. If the polynomial is added to another polynomial, the resulting expression is also a polynomial. The same goes with the operations of addition, subtraction, multiplication and division.

In this article, we will see how to find the unknown constants, and how to multiply and divide the polynomials. Below are some of the examples of polynomial word problems which you will find quite useful in understanding polynomials and their attributes when they are added, subtracted, multiplied or divided.

x^ 5 - ax + b

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x^5 - ax + b

Set the factors equal to zero:

(x+2) =0

Add both expressions together to get

2a + b - 32 - 2a + b + 32 = 0

Take 4 on the left side of the equation:

3 + m = -4

Subtract 3 from both sides of the equation to get the final answer:

m = -7

The area of the rectangle = length x width

(2x - 3)(x - 1)

We know that the amount of revenue generated is equal to the:

Number of items sold x Price per item

Multiply these two expressions together:

(7t + 25) \cdot (3 t ^2 +3t +36)

The formula for area of the rectangle = length x width

Hence, to find the width of the rectangle, we need to divide the area by the length:

\frac {x ^ 3 - 2 x ^ 2 -6x + 12}{x - 2}

Use the polynomial long division method to solve the above expression:

\begin{array}{ccccccccccc} &&& x^2 & + & 6 &\\ \cline{3-10} \multicolumn{2}{r}{x - 2 \surd} & x^3 &-& 2x^2 &-& 6x &+& 12 & \\ & &x^3&-& 2x ^ 2&& & & & \\ \cline{3-7} & & & & &-& 6x&+& 12 & & & & & \\ & & & & & -& 6x&+& 12& & && & \\ \cline{4-8} & & & && && 0 \\ \end{array}

Since the formula for the distance is speed x time, hence we can easily derive formula of speed from this formula of distance:

\frac {distance}{time}

Put the values in the questions in the above formula to get the speed:

\frac {2x ^2 +6x -20}{x - 2}

Use the polynomial long division method to find the answer.

\begin{array}{ccccccccccc} &&& 2x & + & 10 & \\ \cline{3-10} \multicolumn{2}{r}{x - 2 \surd} & 2x^2 &+& 6x &-& 20 & \\ && 2x^2&-& 4x && & & & \\ \cline{3-7} & & & & 10x&-& 20& & & & & \\ & & & & 10x&-& 20& & && & \\ \cline{4-8} & & & && && 0 \\ \end{array}

The total amount of profit is calculated by the formula:

Profit = Price per item x Number of items sold

Hence, we will find the profit by multiplying the price of the single shirt with the total number of shirts sold.

(2x + 1)(3x - 5)

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Dividing Algebraic Fractions

Operations with monomials, algebraic identities, polynomial roots, factoring polynomials, dividing polynomials, algebraic fractions, adding or subtracting algebraic fractions, binomial theorem, adding polynomials, reducing algebraic fractions to a common denominator, multiplying polynomials, algebraic expressions, remainder theorem, multiplying algebraic fractions, binimial theorem worksheet, polynomials, ruffini’s rule, binomial formula, polynomial formulas, algebra formulas, algebraic fractions worksheet, polynomial worksheet, factoring polynomials worksheet, binomial worksheet, monomial worksheets, cancel reply.

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The area of his garden is x^2 + 3x +2 and its width is x + 1. Using division of polynomials, finds its length.

It is very important for students

There is a mistake I the solution to no 7 the dividing part . It’s supposed to be x²-6

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  6. Solving Polynomial Equations

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  1. What Are the Six Steps of Problem Solving?

    The six steps of problem solving involve problem definition, problem analysis, developing possible solutions, selecting a solution, implementing the solution and evaluating the outcome. Problem solving models are used to address issues that...

  2. What Is the Answer to Your Math Problem?

    The answer to any math problem depends on upon the question being asked. In most math problems, one needs to determine a missing variable. For instance, if a problem reads as 2+3 = , one needs to figure out what the number after the equals ...

  3. How Do You Do Math Percentage Problems?

    To calculate percentages, convert the percentage to a decimal and multiply it by the number in the problem. For example, to find 40 percent of 50, change it to 0.40 times 50, which gives you the result of 20.

  4. Solve

    For example, 7 x 3 y 2 z 5 is a monomial in the variables x , y , and z . Constants are also referred to as monomials. A polynomial in the variables x , y , …

  5. Algebra

    Section 1.4 : Polynomials · Add 4x3−2x2+1 4 x 3 − 2 x 2 + 1 to 7x2+12x 7 x 2 + 12 x Solution · Subtract 4z6−3z2+2z 4 z 6 − 3 z 2 + 2 z from −

  6. Polynomial Equation Calculator

    To solve a polynomial equation write it in standard form (variables and canstants on one side and zero on the other side of the equation). Factor it and set

  7. Solving Polynomial Equations

    For a complete lesson on solving polynomial equations, go to https://www.MathHelp.com - 1000+ online math lessons featuring a personal math

  8. Solving Polynomials

    Read how to solve Quadratic Polynomials (Degree 2) with a little work,; It can be hard to solve Cubic (degree 3) and Quartic (degree 4) equations,; And beyond

  9. 4.4 Solve Polynomial Equations by Factoring

    To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The

  10. Polynomial equation solver

    This online calculator provides a step-by-step explanation of how to solve polynomial equations.

  11. Polynomials

    Algebraic expressions and polynomials. Calculate the sum, difference, product and quotient of polynomials and algebraic expressions on Math-Exercises.com.

  12. How to Solve Polynomials: 13 Steps (with Pictures)

    If you need to solve a quadratic polynomial, write the equation in order of the highest degree to the lowest, then set the equation to equal zero. Rewrite the

  13. Solving Polynomials: How-to

    Note: The terminology for this topic is often used carelessly. Technically, one "solves" an equation, such as "(polynomal) equals (zero)"; one "finds the roots"

  14. Polynomial Word Problems

    Solution of exercise Solved Polynomial Word Problems ... Calculate the value of a for which the polynomial x ^ 3 - ax + 8