# How do you factor  4x^3 - x^2 -12x + 3 by grouping?

Jun 23, 2016

$\left(4 x - 1\right) \left({x}^{2} - 3\right)$

#### Explanation:

Think of this cubic as two groups:

${\overbrace{\left(4 {x}^{3} - {x}^{2}\right)}}^{\text{Group 1"+overbrace((-12x+3))^"Group 2}}$

We will want to find a common factor in each group. From $4 {x}^{3} - {x}^{2}$, we see a common factor of ${x}^{2}$:

${\overbrace{{x}^{2} \left(4 x - 1\right)}}^{\text{Group 1"+overbrace((-12x+3))^"Group 2}}$

From Group 2, we could either factor out a $+ 3$ or a $- 3$. To make the leading term positive, we will factor out a $- 3$ from both $- 12 x$ and $3$, leaving:

${\overbrace{{x}^{2} \left(4 x - 1\right)}}^{\text{Group 1"+overbrace(-3(4x-1))^"Group 2}}$

From here, notice that there is a common factor between Group 1 and Group 2. Both have terms, ${x}^{2}$ and $- 3$, which are being multiplied by the other term $\left(4 x - 1\right)$. Here, $\left(4 x - 1\right)$, despite it being made up of two terms, is the common factor and can be factored out. The ${x}^{2}$ and $- 3$ are combined since they share this common factor:

$\left(4 x - 1\right) \left({x}^{2} - 3\right)$

Depending on your level of instruction, you may recognize that ${x}^{2} - 3$ is a difference of squares, and can be factorized yet. If not, this is a fine final answer.