To review these concepts, go to Polynomial Division Help and Synthetic Division Help
Polynomial Division Practice Problems
Problems
For Problems 1-4 use long division to find the quotient and remainder. For Problems 5 and 6, use synthetic division
-
(6 x 3 − 2 x 2 + 5 x − 1) ÷ ( x 2 + 3 x + 2)
-
( x 3 − x 2 + 2 x + 5) ÷ (3 x − 4)

-
( x 3 + 8) ÷ ( x + 2)
-
Use synthetic division and the Remainder Theorem to evaluate f(c) .
f(x) = 6 x 4 − 8 x 3 + x 2 + 2 x − 5; c = −2
-
Completely factor the polynomial. f(x) = x 3 + 2 x 2 − x − 2; c = 1 is a zero.
-
Completely factor the polynomial. P(x) = x 3 − 5 x 2 + 5 x + 3; c = 3 is a zero.
Solutions
-

The quotient is 6 x − 20, and the remainder is 53 x + 39.
-

-


The quotient is −6 x + 2, and the remainder is 10 x .
-

The quotient is x 2 + x + 1, and the remainder is 0.
-

The quotient is x 2 − x + 4, and the remainder is −20.
-

The quotient is x 2 − 2 x + 4, and the remainder is 0.
-

The remainder is 155, so f (−2) = 155.
-

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

P(x) = ( x − 3)( x 2 − 2 x − 1)
In order to factor x 2 − 2 x − 1, we must first find its zeros.

Because
is a zero,
=
is a factor.Because
is a zero,
=
is a factor.
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