# How do you factor 7y^2-11y-27?

##### 2 Answers
Jan 9, 2017

$7 {y}^{2} - 11 y - 27 = \frac{1}{28} \left(14 y - 11 - \sqrt{877}\right) \left(14 y - 11 + \sqrt{877}\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Use this with $a = \left(14 y - 11\right)$ and $b = \sqrt{877}$

First premultiply by $7 \cdot {2}^{2} = 28$, then divide by it after completing the square:

$28 \left(7 {y}^{2} - 11 y - 27\right) = 196 {y}^{2} - 308 y - 756$

$\textcolor{w h i t e}{28 \left(7 {y}^{2} - 11 y - 27\right)} = {\left(14 y\right)}^{2} - 2 \left(14 y\right) \left(11\right) + 121 - 877$

$\textcolor{w h i t e}{28 \left(7 {y}^{2} - 11 y - 27\right)} = {\left(14 y - 11\right)}^{2} - {\left(\sqrt{877}\right)}^{2}$

$\textcolor{w h i t e}{28 \left(7 {y}^{2} - 11 y - 27\right)} = \left(\left(14 y - 11\right) - \sqrt{877}\right) \left(\left(14 y - 11\right) + \sqrt{877}\right)$

$\textcolor{w h i t e}{28 \left(7 {y}^{2} - 11 y - 27\right)} = \left(14 y - 11 - \sqrt{877}\right) \left(14 y - 11 + \sqrt{877}\right)$

Hence:

$7 {y}^{2} - 11 y - 27 = \frac{1}{28} \left(14 y - 11 - \sqrt{877}\right) \left(14 y - 11 + \sqrt{877}\right)$

Jan 9, 2017

(1/28)(14x - 11 - sqrt877)(14x - 11 + sqrt877)

#### Explanation:

There is another classical way.
f(x) = a(x - x1)(x - x2),
where x1 and x2 are the 2 real roots of the quadratic equation
f(x) = 7x^2 - 11x - 27 = 0.
$D = {d}^{2} = {b}^{2} - 4 a c = 121 + 756 = 877$ --> $d = \pm \sqrt{877}$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{11}{14} \pm \frac{\sqrt{877}}{14}$
$x 1 = \frac{11}{14} + \frac{\sqrt{877}}{14}$
$x 2 = \frac{11}{14} - \frac{\sqrt{877}}{14}$
The factored form will be:
$f \left(x\right) = \left(7\right) \left(x - \frac{11}{14} - \frac{\sqrt{877}}{14}\right) \left(x - \frac{11}{14} - \frac{\sqrt{877}}{14}\right) =$
$= \frac{1}{28} \left(14 x - 11 - \sqrt{877}\right) \left(14 x - 11 + \sqrt{877}\right)$