# How do you factor 21n ^ { 2} + 3n - 24?

Jun 21, 2018

$3 \left(n - 1\right) \left(7 n + 8\right)$

#### Explanation:

$\text{take out a "color(blue)"common factor } 3$

$= 3 \left(7 {n}^{2} + n - 8\right)$

$\text{factor the quadratic using the a-c method}$

$\text{the factors of the product } 7 \times - 8 = - 56$

$\text{which sum to "+1" are "-7" and } + 8$

$\text{split the middle term using these factors}$

$7 {n}^{2} - 7 n + 8 n - 8 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$= \textcolor{red}{7 n} \left(n - 1\right) \textcolor{red}{+ 8} \left(n - 1\right)$

$\text{take out the "color(blue)"common factor } \left(n - 1\right)$

$= \left(n - 1\right) \left(\textcolor{red}{7 n + 8}\right)$

$21 {n}^{2} + 3 n - 24 = 3 \left(n - 1\right) \left(7 n + 8\right)$

Jun 21, 2018

$21 \cdot \left(n - 1\right) \left(n + \frac{8}{7}\right)$

#### Explanation:

Solving the equation$21 {n}^{2} + 3 n - 24 = 0$
by the quadratic Formula we get

${n}_{1 , 2} = - \frac{1}{14} \pm \frac{15}{14}$
so
${n}_{1} = 1$
${n}_{2} = - \frac{8}{7}$
and we get
$21 \left(n - 1\right) \left(n + \frac{8}{7}\right)$