Question

1. which expressions are equivalent to $6x^2 + 3x - 9$?

select two correct answers.
$\square$ $3(2x + 3)(x - 1)$
$\square$ $-3(2x^2 + x + 3)$
$\square$ $3(2x^2 + x - 3)$
$\square$ $-3(2x - 3)(x - 1)$
$\square$ $3(2x - 3)(x + 1)$
clear all

Explanation:

Step1: Analyze the first option

Expand \(3(2x + 3)(x - 1)\):
First, multiply \((2x + 3)(x - 1)\) using the distributive property (FOIL method):
\((2x + 3)(x - 1)=2x\times x-2x\times1 + 3\times x-3\times1=2x^{2}-2x + 3x-3=2x^{2}+x - 3\)
Then multiply by 3: \(3(2x^{2}+x - 3)=6x^{2}+3x - 9\), so this option is equivalent.

Step2: Analyze the second option

Expand \(-3(2x^{2}+x + 3)\):
Using the distributive property: \(-3\times2x^{2}-3\times x-3\times3=-6x^{2}-3x - 9\), which is not equivalent to \(6x^{2}+3x - 9\).

Step3: Analyze the third option

Expand \(3(2x^{2}+x - 3)\):
Using the distributive property: \(3\times2x^{2}+3\times x-3\times3 = 6x^{2}+3x - 9\), so this option is equivalent.

Step4: Analyze the fourth option

Expand \(-3(2x - 3)(x - 1)\):
First, multiply \((2x - 3)(x - 1)\): \(2x\times x-2x\times1-3\times x + 3\times1=2x^{2}-2x-3x + 3=2x^{2}-5x + 3\)
Then multiply by - 3: \(-3(2x^{2}-5x + 3)=-6x^{2}+15x - 9\), not equivalent.

Step5: Analyze the fifth option

Expand \(3(2x - 3)(x + 1)\):
First, multiply \((2x - 3)(x + 1)\): \(2x\times x+2x\times1-3\times x-3\times1=2x^{2}+2x-3x - 1=2x^{2}-x - 1\)
Then multiply by 3: \(3(2x^{2}-x - 1)=6x^{2}-3x - 3\), not equivalent.

Answer:

A. \(3(2x + 3)(x - 1)\)
C. \(3(2x^{2}+x - 3)\)