How do you factor 2x^2+15x+18?

Jun 16, 2016

$\left(2 x + 3\right) \left(x + 6\right)$

Explanation:

$2 {x}^{2} + 15 x + 18 = 2 {x}^{2} + 12 x + 3 x + 18 = 2 x \left(x + 6\right) + 3 \left(x + 6\right) = \left(2 x + 3\right) \left(x + 6\right)$[Ans]

Jun 16, 2016

$\textcolor{m a \ge n t a}{2} {x}^{2} + 15 x \textcolor{\lim e}{+} \textcolor{p u r p \le}{18}$

Find factors of $2 \mathmr{and} 18$ which ADD (because of $\textcolor{\lim e}{+}$ 18) to make $15.$

The signs will be the same (because of the color(lime)(+)18), they are both positive (because of the +15).

$2 \text{ } 18$
$\downarrow \text{ } \downarrow$

$\textcolor{red}{2} \text{ "color(blue)(3) " cross multiply and ADD the products}$
$\textcolor{b l u e}{1} \text{ } \textcolor{red}{6}$

$\textcolor{red}{2} \text{ } \textcolor{b l u e}{3 \Rightarrow 1 \times 3 = 3}$
 color(blue)(1)" "color(red)(6 rArr 2 xx6 = ul12
$\textcolor{w h i t e}{\times \times \times \times . \times \times x} 15$

The top line gives one of the factors and the bottom line gives the other factor.

$\left(2 x + 3\right) \left(x + 6\right)$

Jun 16, 2016

$\left(2 x + 3\right) \left(x + 6\right)$

Explanation:

Use the new AC method (Socratic Search)

$y = 2 {x}^{2} + 15 x + 18 = 2 \left(x + p\right) \left(x + q\right)$

Converted trinomial: $y ' = {x}^{2} + 15 x + 36 = \left(x + p '\right) \left(x + q '\right) .$

Find $p ' \mathmr{and} q '$ knowing sum $\left(b = 15\right)$ and product $\left(a c = 36\right) .$

They are: $p ' = 3 \mathmr{and} q ' = 12.$
Back to original $y \rightarrow p = \frac{p '}{a} = \frac{3}{2}$, and $q = \frac{q '}{a} = \frac{12}{2} = 6$.

Factored form: $2 \left(x + \frac{3}{2}\right) \left(x + 6\right) = \left(2 x + 3\right) \left(x + 6\right)$