Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Chapter Review
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Exercise 34 Page 349

Use synthetic division to divide the polynomial by the binomial. If the remainder is zero, the binomial is a factor of a polynomial.

No.

Practice makes perfect
To divide polynomials using synthetic division, all the terms of the dividend must be present. Since there are no missing terms, we do not need to rewrite the polynomial. x^3+x^2-10x+8 Remember that the general form of the divisor in synthetic division is x-a. Since the given divisor is already written in this form, we are ready to go!
rl IR-0.15cm r 4 & |rr 1 &1 &-10 & 8

Bring down the first coefficient

rl IR-0.15cm r 4 & |rr 1 &1 &-10 &8 &&& & c 1 & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 4 & |rr 1 &1 &-10 &8 &4&& & c 1 & & &

Add down

rl IR-0.15cm r 4 & |rr 1 &1 &-10 &8 &4&& & c 1 &5 &
â–Ľ
Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r 4 & |rr 1 &1 &-10 &8 &4& 20& & c 1 & 5 & &

Add down

rl IR-0.15cm r 4 & |rr 1 &1 &-10 &8 &4& 20& & c 1 &5 & 10 &

Multiply the coefficient by the divisor

rl IR-0.15cm r 4 & |rr 1 &1 &-10 &8 &4& 20&40 & c 1 &5 & 10 &

Add down

rl IR-0.15cm r 4 & |rr 1 &1 &-10 &8 &4& 20&40 & c 1 &5 & 10 &48
The remainder is not zero , so x-4 is not a factor of x^3+x^2-10x+8.