# How do you factor 5x^{2} + 11x + 7?

Use the discriminant to determine whether or not there are real roots, then use the quadratic formula to solve for those roots

#### Explanation:

The discriminant is the part of the quadratic formula under the square root

${b}^{2} - 4 a c$

sub in the terms from the standard form of the equation
ex. $a {x}^{2} + b x + c$

so in this case $a = 5 , b = 11 , c = 7$

If the discriminant is positive there are 2 real roots, if the discriminant equals 0 there is one real root, and if the discriminant is negative there are no real roots

${11}^{2} - 4 \left(5\right) \left(7\right)$
$= 121 - 140$
$= - 19$

Since the discriminant is negative, there are no real roots and you cannot factor this equation

Oct 14, 2016

$5 {x}^{2} + 11 x + 7 = \frac{1}{20} \left(10 x + 11 - \sqrt{19} i\right) \left(10 x + 11 + \sqrt{19} i\right)$

#### Explanation:

This quadratic has negative discriminant, so can only be factored using Complex coefficients.

$\Delta = {b}^{2} - 4 a c = {11}^{2} - 4 \left(5\right) \left(7\right) = 121 - 140 = - 19$

Multiply by $20 = {2}^{2} \cdot 5$, complete the square, use the difference of squares identity, then divide by $20$...

The difference of squares identity can be written:

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

We use this with $a = 10 x + 11$ and $b = \sqrt{19} i$ as follows:

$20 \left(5 {x}^{2} + 11 x + 7\right) = 100 {x}^{2} + 220 x + 140$

$\textcolor{w h i t e}{20 \left(5 {x}^{2} + 11 x + 7\right)} = {\left(10 x\right)}^{2} + 2 \left(11\right) \left(10 x\right) + 121 + 19$

$\textcolor{w h i t e}{20 \left(5 {x}^{2} + 11 x + 7\right)} = {\left(10 x + 11\right)}^{2} + 19$

$\textcolor{w h i t e}{20 \left(5 {x}^{2} + 11 x + 7\right)} = {\left(10 x + 11\right)}^{2} - {\left(\sqrt{19} i\right)}^{2}$

$\textcolor{w h i t e}{20 \left(5 {x}^{2} + 11 x + 7\right)} = \left(\left(10 x + 11\right) - \sqrt{19} i\right) \left(\left(10 x + 11\right) + \sqrt{19} i\right)$

$\textcolor{w h i t e}{20 \left(5 {x}^{2} + 11 x + 7\right)} = \left(10 x + 11 - \sqrt{19} i\right) \left(10 x + 11 + \sqrt{19} i\right)$

So:

$5 {x}^{2} + 11 x + 7 = \frac{1}{20} \left(10 x + 11 - \sqrt{19} i\right) \left(10 x + 11 + \sqrt{19} i\right)$