To review these concepts, go to Polynomial Division Help and Synthetic Division Help
Polynomial Division Practice Problems
Problems
For Problems 14 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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