Dividing a Polynomial by a Binomial
We can multiply x - 2 and x + 5 to get
(x - 2)(x + 5) = x2 + 3x - 10.
So if we divide x2 + 3x - 10 by the factor x - 2, we should get the other factor
x + 5. This division is not done like division by a monomial; it is done like long
division of whole numbers. We get the first term of the quotient by dividing the
first term of x - 2 into the first term of x2 + 3x - 10. Divide x2 by x to get x.
Now bring down -10. We get the second term of the quotient (below) by dividing
the first term of x - 2 into the first term of 5x - 10. Divide 5x by x to get 5.
So the quotient is x + 5 and the remainder is 0. If the remainder is not 0, then
dividend = (divisor)(quotient) + (remainder).
If we divide each side of this equation by the divisor, we get
||= quotient +
When dividing polynomials, we must write the terms of the divisor and the dividend
in descending order of the exponents. If any terms are missing, as in the next
example, we insert terms with a coefficient of 0 as placeholders. When dividing
polynomials, we stop the process when the degree of the remainder is smaller than
the degree of the divisor.
Find the quotient and remainder for (3x4 - 2 - 5x) Ã· (x2
Rearrange 3x4 - 2 - 5x as 3x4 - 5x - 2 and insert the terms 0x3 and 0x2:
Students usually have the
most difficulty with the subtraction
part of long division.
So pay particular attention to
that step and double check
The quotient is 3x2 + 9x + 27, and the remainder is 76x - 2. Note that the degree
of the remainder is 1, and the degree of the divisor is 2. To check, verify that
(x2 - 3x)(3x2 + 9x + 27) + 76x - 2 = 3x4 - 5x
Rewriting a ratio of two polynomials
in the form
Divide 4x3 - x - 9 by 2x 3. Insert 0 Â· x2 for the missing term.
Since the quotient is 2x2 + 3x + 4 and the remainder is 3, we have
4x3 - x - 9
|= 2x2 + 3x + 4 +
2x - 3
2x - 3
To check the answer, we must verify that
(2x - 3)(2x2 + 3x + 4) + 3 = 4x3 - x - 9.