Try the Free Math Solver or Scroll down to Tutorials!

 Depdendent Variable

 Number of equations to solve: 23456789
 Equ. #1:
 Equ. #2:

 Equ. #3:

 Equ. #4:

 Equ. #5:

 Equ. #6:

 Equ. #7:

 Equ. #8:

 Equ. #9:

 Solve for:

 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
 Ineq. #2:

 Ineq. #3:

 Ineq. #4:

 Ineq. #5:

 Ineq. #6:

 Ineq. #7:

 Ineq. #8:

 Ineq. #9:

 Solve for:

 Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:

# Dividing Polynomials by Monomials and Binomials

## Dividing a Polynomial by a Monomial

We know how to divide some simple polynomials by monomials. For example, We use the distributive property to take one-half of 6x and one-half of 8 to get 3x + 4. So both 6x and 8 are divided by 2. To divide any polynomial by a monomial, we divide each term of the polynomial by the monomial.

Example 1

Dividing a polynomial by a monomial

Find the quotient for (-8x6 + 12x4 - 4x2) Ã· (4x2).

Solution The quotient is -2x4 + 3x2 - 1. We can check by multiplying.

4x2(-2x4 + 3x2 - 1) = -8x6 + 12x4 - 4x2.

Because division by zero is undefined, we will always assume that the divisor is nonzero in any quotient involving variables. For example, the division in Example 1 is valid only if 4x2 ≠ 0, or x2 ≠ 0.

## Dividing a Polynomial by a Binomial

Division of whole numbers is often done with a procedure called long division. For example, 253 is divided by 7 as folows: Note that 36 Â· 7 + 1 = 253. It is always true that

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

To divide a polynomial by a binomial, we perform the division like long division of whole numbers. For example, to divide x2 - 3x - 10 by x + 2, we get the first term of the quotient by dividing the first term of x + 2 into the first term of x2 - 3x - 10. So divide x2 by x to get x, then multiply and subtract as follows:

 1) Divide:2) Multiply: 3) Subtract: x2 Ã· x = x x Â· (x + 2) = x2 + 2x -3x - 2x = -5x

Now bring down -10 and continue the process. We get the second term of the quotient (below) by dividing the first term of x + 2 into the first term of -5x - 10. So divide -5x by x to get -5:

 1) Divide:2) Multiply:   3) Subtract: -5x Ã· x = -5 Bring down -10. -5(x + 2) = -5x - 10 -10 - (-10) = 0

So the quotient is x - 5, and the remainder is 0.

In the next example we must rearrange the dividend before dividing.

Example 2

Dividing a polynomial by a binomial

Divide 2x3 - 4 - 7x2 by 2x - 3, and identify the quotient and the remainder.

Solution

Rearrange the dividend as 2x3 - 7x2  - 4. because the x-term in the dividend is missing, we write 0 Â· x for it: The quotient is x2 - 2x - 3, and the remainder is -13. Note that the degree of the remainder is 0 and the degree of the divisor is 1. To check, we must verify that

(2x - 3)(x2 - 2x - 3)  - 13 = 2x3 - 7x2  - 4.

Caution

To avoid errors, always write  the terms of the divisor and the dividend in descending order of the exponents and insert a zero for any term that is missing.

If we divide both sides of the equation

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

by te divisor, we get the equation This fact is used in expressing improper fractions as mixed numbers. For example, if 19 is divided by 5, the quotient is 3 and the remainder is 4. So We can also use this form to rewrite algebraic fractions.