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 onehalf of 6x and onehalf 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 (8x^{6} + 12x^{4}  4x^{2})
Ã· (4x^{2}).
Solution
The quotient is 2x^{4} + 3x^{2}  1. We can check by
multiplying.
4x^{2}(2x^{4} + 3x^{2}  1) = 8x^{6} + 12x^{4}
 4x^{2}.
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 4x^{2} ≠ 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 x^{2}  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 x^{2}  3x  10. So divide x^{2} by x to
get x, then multiply and subtract as follows:
1) Divide: 2) Multiply:
3) Subtract: 

x^{2} Ã· x = x
x Â· (x + 2) = x^{2} + 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 2x^{3}  4  7x^{2} by 2x  3, and identify the
quotient and the remainder.
Solution
Rearrange the dividend as 2x^{3}  7x^{2}  4. because
the xterm in the dividend is missing, we write 0 Â· x for it:
The quotient is x^{2}  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)(x^{2}  2x  3)  13 = 2x^{3}  7x^{2}
 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.