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# Polynomial Division Help

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By McGraw-Hill Professional
Updated on Oct 4, 2011

## Introduction to Polynomial Division

Polynomials can be divided in much the same way as whole numbers. When we take the quotient of two whole numbers (where the divisor is not zero), we get a quotient and a remainder. The same happens when we take the quotient of two polynomials. Polynomial division is useful when factoring polynomials.

Polynomial division problems usually come in one of two forms.

According to the division algorithm for polynomials, for any polynomials f(x) and g(x) (with g(x) not the zero function)

where q(x) is the quotient (which might be 0) and r(x) is the remainder, which has degree strictly less than the degree of g(x) . Multiplying by g(x) to clear the fraction, we also get f(x) = g(x)q(x) + r(x) . First we will perform polynomial division using long division.

#### Examples

Find the quotient and remainder using long division.

• We will begin by dividing the leading term of the dividend by the leading term of the divisor. For the first step in this example, we will divide 4 x 2 by x. You might see right away that 4 x 2 ÷ x is 4 x. If not, write 4 x 2 ÷ x as a fraction then reduce: . This will be the first term of the quotient.

Multiply 4 x by the divisor: 4 x (x + 2) = 4 x 2 + 8 x . Subtract this from the first two terms of the dividend. Be careful to subtract all of 4 x 2 + 8 x , not just 4 x 2.

Bring down the next term.

Start the process again with −5 x ÷ x = −5. The next term in the quotient is −5. Multiply x + 2 by −5: −5( x + 2) = −5 x − 10. Subtract this from -5 x − 5.

We are done because cannot be a term in a polynomial. The remainder is 5 and the quotient is 4 x − 5.

• Divide 3 x 4 by x 2 to get the first term of the quotient: . Multiply x 2 + 2 x − 3 by 3 x 2 : 3 x 2 ( x 2 + 2 x − 3) = 3 x 4 + 6 x 3 − 9 x 2 . Subtract this from the first three terms in the dividend.

Divide − x 3 by x 2 to get the second term in the quotient: . Multiply x 2 + 2 x − 3 by − x : − x ( x 2 + 2 x − 3) = − x 3 − 2 x 2 + 3 x. Subtract this from − x 3 + 5 x 2 + 7 x.

Divide 7 x 2 by x 2 to get the third term in the quotient: . Multiply x 2 + 2 x − 3 by 7: 7( x 2 + 2 x − 3) = 7 x 2 + 14 x − 21. Subtract this from 7 x 2 + 4 x − 1.

Because cannot be a term in a polynomial, we are done. The quotient is 3 x 2x + 7, and the remainder is −10 x + 20.

It is important that every power of x , from the highest power to the constant term, be represented in the polynomial. Although it is possible to perform long division without all powers represented, it is very easy to make an error. Also, it is not possible to perform synthetic division (later in this chapter) without a coefficient for every term. If a power of x is not written, we need to rewrite the polynomial (either the dividend, divisor, or both) using a coefficient of 0 on the missing powers. For example, we would write x 3 − 1 as x 3 + 0 x 2 + 0 x − 1.

#### Example

• ( x 3 − 8) ÷ ( x + 1)
• Rewrite as ( x 3 + 0 x 2 + 0 x − 8) ÷ ( x + 1)

The quotient is x 2x + 1, and the remainder is −9.

Polynomial division is a little trickier when the leading coefficient of the divisor is not 1. The terms of the quotient are harder to find and are likely to be fractions.

#### Examples

Find the quotient and remainder using long division.

• Find the first term in the quotient by dividing the first term of the dividend by the first term in the divisor:

The second term in the quotient is

• Find the first term in the quotient by dividing the leading term in the quotient by the first term in the divisor.

The quotient is , and the remainder is .

Practice Problems for this concept can be found at: Polynomial Division Practice Problems

Additional Practice problems for this concept can be found at: Polynomial Functions Practice Test.

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