Question

which shows one way to determine the factors of $12x^3 - 2x^2 + 18x - 3$ by grouping?
$\bigcirc\\ 2x^2(6x - 1) + 3(6x - 1)$
$\bigcirc\\ 2x^2(6x - 1) - 3(6x - 1)$
$\bigcirc\\ 6x(2x^2 - 3) - 1(2x^2 - 3)$
$\bigcirc\\ 6x(2x^2 + 3) + 1(2x^2 + 3)$

Explanation:

Step1: Group the terms

Group the first two terms and the last two terms of the polynomial \(12x^{3}-2x^{2}+18x - 3\). So we have \((12x^{3}-2x^{2})+(18x - 3)\).

Step2: Factor out GCF from each group

For the first group \(12x^{3}-2x^{2}\), the greatest common factor (GCF) is \(2x^{2}\). Factoring out \(2x^{2}\), we get \(2x^{2}(6x - 1)\).

For the second group \(18x - 3\), the GCF is \(3\). Factoring out \(3\), we get \(3(6x - 1)\) (since \(18x\div3 = 6x\) and \(- 3\div3=-1\), wait, no: \(18x-3 = 3\times6x-3\times1=3(6x - 1)\)? Wait, no, \(18x-3=3(6x - 1)\)? Wait, \(3\times6x = 18x\) and \(3\times(- 1)=-3\), yes. Wait, but the original polynomial is \(12x^{3}-2x^{2}+18x - 3\), so when we factor the second group, it's \(+3(6x - 1)\)? Wait, no, let's check the options. Wait, the first option is \(2x^{2}(6x - 1)+3(6x - 1)\), let's expand this: \(2x^{2}\times6x-2x^{2}\times1 + 3\times6x-3\times1=12x^{3}-2x^{2}+18x - 3\), which matches the original polynomial. Let's check other options:

Second option: \(2x^{2}(6x - 1)-3(6x - 1)=12x^{3}-2x^{2}-18x + 3\), which is not the original.

Third option: \(6x(2x^{2}-3)-1(2x^{2}-3)=12x^{3}-18x-2x^{2}+3=12x^{3}-2x^{2}-18x + 3\), not original.

Fourth option: \(6x(2x^{2}+3)+1(2x^{2}+3)=12x^{3}+18x+2x^{2}+3=12x^{3}+2x^{2}+18x + 3\), not original.

So the first option is correct.

Answer:

\(2x^{2}(6x - 1)+3(6x - 1)\) (the first option)