# How do you factor the trinomial x^2+3x-36=0?

Dec 26, 2015

${x}^{2} + 3 x - 36 = 0$ can be solved by quadratic formula but cannot be solved by integer factors. In case, you are looking for a solution it is explained below.

#### Explanation:

Quadratic equations of for $a {x}^{2} + b x + c = 0$ can be solved by the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Our question ${x}^{2} + 3 x - 36 = 0$
Compare with $a {x}^{2} + b x + c = 0$

$a = 1$, $b = 3$ and $c = - 36$

Substitute these values of $a$,$b$ and $c$ in the formula

$x = \frac{- \left(3\right) \pm \sqrt{{\left(3\right)}^{2} - 4 \left(1\right) \left(- 36\right)}}{2 \left(1\right)}$

$x = \frac{- 3 \pm \sqrt{9 + 144}}{2}$

$x = \frac{- 3 \pm \sqrt{153}}{2}$

$x = \frac{- 3 \pm \sqrt{{3}^{2} \cdot 17}}{2}$
$x = \frac{- 3 \pm 3 \sqrt{17}}{2}$

Solutions:

$x = \frac{- 3 - 3 \sqrt{17}}{2}$ or $x = \frac{- 3 + 3 \sqrt{17}}{2}$

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