Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
4. Dividing Polynomials
Continue to next subchapter

Exercise 30 Page 308

Rewrite the given divisor into the general form x-a. Then, if all the terms of the dividend are present, you are ready to use synthetic division.

(x+3)(x−3)(x−4)

Practice makes perfect
To divide the given 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 - 4x^2 - 9x + 36 Remember that the general form of the divisor in synthetic division is x-a. Therefore, we need to rewrite ours a little bit. x+3 ⇔ x-(-3) Now, we are ready to divide!
rl IR-0.15cm r - 3 & |rr 1& - 4 & - 9 &36

Bring down the first coefficient

rl IR-0.15cm r - 3 & |rr 1 & - 4 & - 9 & 36 & c 1 & & &

Multiply the coefficient by the divisor

rl IR-0.15cm r - 3 & |rr 1& - 4 & - 9 & 36 &- 3&& & c 1& & &

Add down

rl IR-0.15cm r - 3 & |rr 1& - 4 & - 9 & 36 &- 3&& & c 1 & - 7 & &
Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r - 3 & |rr 1& - 4 & - 9 & 36 &- 3&21& & c 1 & -7 & &

Add down

rl IR-0.15cm r - 3 & |rr 1& - 4 & - 9 & 36 &- 3&21& & c 1 & -7 &12&

Multiply the coefficient by the divisor

rl IR-0.15cm r - 3 & |rr 1& - 4 & - 9 & 36 &- 3&21&- 36 & c 1 & -7 & 12&

Add down

rl IR-0.15cm r - 3 & |rr 1& - 4 & - 9 & 36 &- 3&21&- 36 & c 1 & -7 &12& 0
The quotient is a polynomial of degree 2 and the remainder is zero. We know that our remainder is correct because we were told that x+3 is a factor of the polynomial. Let's rewrite the given polynomial as the product of two factors. ( x^3 - 4x^2 - 9x + 36 ) = (x+3) ( x^2 - 7x + 12 ) Finally, let's factor the quadratic polynomial. ( x^3 - 4x^2 - 9x + 36 ) = (x+3)(x-3)(x-4)