# How do you factor the trinomial x^2 +3x-36?

Jan 29, 2018

${x}^{2} + 3 x - 36 = \left(x + \frac{3}{2} - \frac{3}{2} \sqrt{17}\right) \left(x + \frac{3}{2} + \frac{3}{2} \sqrt{17}\right)$

#### Explanation:

We can factor this trinomial by completing the square. To avoid some of the arithmetic involving fractions, multiply by $4$ first...

$4 \left({x}^{2} + 3 x - 36\right) = 4 {x}^{2} + 12 x - 144$

$\textcolor{w h i t e}{4 \left({x}^{2} + 3 x - 36\right)} = 4 {x}^{2} + 12 x + 9 - 153$

$\textcolor{w h i t e}{4 \left({x}^{2} + 3 x - 36\right)} = {\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(3\right) + {3}^{2} - 153$

$\textcolor{w h i t e}{4 \left({x}^{2} + 3 x - 36\right)} = {\left(2 x + 3\right)}^{2} - {\left(3 \sqrt{17}\right)}^{2}$

$\textcolor{w h i t e}{4 \left({x}^{2} + 3 x - 36\right)} = \left(\left(2 x + 3\right) - 3 \sqrt{17}\right) \left(\left(2 x + 3\right) + 3 \sqrt{17}\right)$

$\textcolor{w h i t e}{4 \left({x}^{2} + 3 x - 36\right)} = \left(2 x + 3 - 3 \sqrt{17}\right) \left(2 x + 3 + 3 \sqrt{17}\right)$

So:

${x}^{2} + 3 x - 36 = \left(x + \frac{3}{2} - \frac{3}{2} \sqrt{17}\right) \left(x + \frac{3}{2} + \frac{3}{2} \sqrt{17}\right)$