Divide Rational Expressions...2

[nycmathguy]

College Algebra
Chapter 1/Section 7

 

[HallsofIvy]

Yes. Assuming you want to factor with integer coefficients then the factors of 9x^2+ 3x- 2 must have constant term 1, -2 or -1, 2. The coefficient of x must be 9, 1 or 3, 3 or -9,-1 or -3, -3.
So possible factors are
(9x+ 1)(x- 2)= 9x^2- 17x- 2
(9x- 1)(x+ 2)= 9x^2+ 17xx- 2
(x+ 1)(9x- 2)= 9x^2+ 7x- 2
(x- 1)(9x+ 2)= 9x^2- 7x- 2
(3x+ 1)(3x- 2)= 9x^2- 3x- 2
(3x- 1)(3x+ 2)= 9x^2+ 3x- 2

So the correct factorization is (3x- 1)(3x+ 2).

 

[nycmathguy]

Yes. Assuming you want to factor with integer coefficients then the factors of 9x^2+ 3x- 2 must have constant term 1, -2 or -1, 2. The coefficient of x must be 9, 1 or 3, 3 or -9,-1 or -3, -3.
So possible factors are
(9x+ 1)(x- 2)= 9x^2- 17x- 2
(9x- 1)(x+ 2)= 9x^2+ 17xx- 2
(x+ 1)(9x- 2)= 9x^2+ 7x- 2
(x- 1)(9x+ 2)= 9x^2- 7x- 2
(3x+ 1)(3x- 2)= 9x^2- 3x- 2
(3x- 1)(3x+ 2)= 9x^2+ 3x- 2

So the correct factorization is (3x- 1)(3x+ 2).

Looks good. Thanks again.