# How do you factor 16x^3-3x^2-64x+12?

Oct 2, 2015

$\left(16 x - 3\right) \left(x + 2\right) \left(x - 2\right)$

#### Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} 16 {x}^{3} - 3 {x}^{2} - 64 x + 12$

Note that the ratio $16 : - 64$ is the same as the ratio $- 3 : 12$
and group the terms accordingly:
$\textcolor{w h i t e}{\text{XXX}} = \left(16 {x}^{3} - 64 x\right) - \left(3 {x}^{2} - 12\right)$

$\textcolor{w h i t e}{\text{XXX}} = \left(16 x\right) \left({x}^{2} - 4\right) - \left(3\right) \left({x}^{2} - 4\right)$

$\textcolor{w h i t e}{\text{XXX}} = \left(16 x - 3\right) \left({x}^{2} - 4\right)$

Recognizing the second factor as the difference of squares:
$\textcolor{w h i t e}{\text{XXX}} = \left(16 x - 3\right) \left(x + 2\right) \left(x - 2\right)$

Oct 2, 2015

$16 {x}^{3} - 3 {x}^{2} - 64 x + 12 = \left(x - 2\right) \left(x + 2\right) \left(16 x - 3\right)$

#### Explanation:

Let's assume that factors contain integer numbers and will look like $\left(x - a\right)$. In this case the value of $a$, if substituted as $x$ would evaluate this expression to $0$, that is $a$ would be a solution of an equation
$16 {x}^{3} - 3 {x}^{2} - 64 x + 12 = 0$

Integer solutions to such an equation with integer coefficients must be among the divisors of a free member $12$, that is they are supposed to be equal to $\pm 2$ and $\pm 3$.

Let's check if any of these four candidates is a solution.
Starting with $x = 2$, we get
$16 \cdot {2}^{3} - 3 \cdot {2}^{2} - 64 \cdot 2 + 12 =$
$= 16 \cdot 8 - 3 \cdot 4 - 64 \cdot 2 + 12 = 128 - 12 - 128 + 12 = 0$

Since $x = 2$ is solution, our expression can be represented as
$16 {x}^{3} - 3 {x}^{2} - 64 x + 12 = \left(x - 2\right) \cdot A$
where $A$ we can determine very simply by the following procedure.

To have $\left(x - 2\right)$ as a factor, we transform the original expression in such a way that $\left(x - 2\right)$ will be a factor for every pair of terms. The term $16 {x}^{3}$ needs the term $- 32 {x}^{2}$ to be able to factor out $\left(x - 2\right)$. Then we add whatever is necessary to get the original $- 3 {x}^{2}$ (that is, we add $29 {x}^{2}$) and continue pairing for another factoring of $\left(x - 2\right)$ etc:
$16 {x}^{3} - 3 {x}^{2} - 64 x + 12 =$
$= 16 {x}^{3} - 32 {x}^{2} + 29 {x}^{2} - 58 x - 6 x + 12 =$
$= 16 {x}^{2} \left(x - 2\right) + 29 x \left(x - 2\right) - 6 \left(x - 2\right) =$
$= \left(x - 2\right) \left(16 {x}^{2} + 29 x - 6\right)$

Let's do the same with the quadratic polynomial $16 {x}^{2} + 29 x - 6$.
Again, looking for integer divisors of the free member $6$ to find the value that evaluates this polynomial to $0$. These are $\pm 2$ and $\pm 3$. Obviously, $2$ is not good, but $- 2$ fits the bill:
$16 {\left(- 2\right)}^{2} + 29 \left(- 2\right) - 6 =$
$= 16 \cdot 4 - 29 \cdot 2 - 6 = 64 - 58 - 6 = 0$

Therefore, we can extract $\left(x + 2\right)$ from the quadratic polynomial. Let's use the same technique as above:
$16 {x}^{2} + 29 x - 6 = 16 {x}^{2} + 32 x - 3 x - 6 =$
$= 16 x \left(x + 2\right) - 3 \left(x + 2\right) = \left(x + 2\right) \left(16 x - 3\right)$

Now we have a complete factorization:
$16 {x}^{3} - 3 {x}^{2} - 64 x + 12 = \left(x - 2\right) \left(x + 2\right) \left(16 x - 3\right)$