Question

Factor completely 3bx2 − 9x3 − b + 3x. (b − 3x)(3x2 − 1) (b + 3x)(3x2 + 1) (b + 3x)(3x2 − 1) Prime

Answer 1

$3bx^2-9x^3-b+3x=\boxed{3x^2}\cdot b-\boxed{3x^2}\cdot3x\fbox{-1}\cdot b\fbox{-1}\cdot (-3x)\\\\=\boxed{3x^2}(b-3x)\fbox{-1}(b-3x)=\boxed{(b-3x)(3x^2-1)}\\\\completely:\\use:a^2-b^2=(a-b)(a+b)\to3x^2-1=(x\sqrt3)^2-1^2=(x\sqrt3-1)(x\sqrt3+1)$
$therefore:\\3bx^2-9x^3-b+3x=(b-3x)(3x^2-1)\\=\underline{\underline{(b-3x)(x\sqrt3-1)(x\sqrt3+1)}}$

Answer 2

Answer: (b-3x)(3x^2 - 1)

Step-by-step explanation:

You have the equation needed to be factored: 3bx^2 − 9x^3 − b + 3x

We first check for GCF: there aren’t any since there are two different variables, x and b as well as x and b are not included in all terms

Therefore, we factor by grouping:

3bx^2 − 9x^3 − b + 3x

Rearranging the equation:

(3b^2 - b) (-9x^3 + 3x)

GCF: b GCF: -3x

b(3x^2 - 1) -3x(3x^2 - 1)

Final answer: (b - 3x)(3x^2 - 1)

:)