# Factorize x^2+6x-5?

Mar 10, 2017

#### Explanation:

As we have ${x}^{2} + 6 x$ in the quadratic polynomial, it is akin to ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$. So we should convert the quadratic polynomial to this form. The process is explained below.

${x}^{2} + 6 x - 5$

= ${x}^{2} + 2 \times x \times 3 + {3}^{2} - {3}^{2} - 5$

= ${\left(x + 3\right)}^{2} - 9 - 5$

= ${\left(x + 3\right)}^{2} - 14$

Here it end the requirement of the question. But if what one wants is to factorize it, one can use ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ and proceed as given below.

${\left(x + 3\right)}^{2} - 14$

= ${\left(x + 3\right)}^{2} - {\left(\sqrt{14}\right)}^{2}$

= $\left(x + 3 + \sqrt{14}\right) \left(x + 3 - \sqrt{14}\right)$

Note: Had it been ${x}^{2} - 6 x$ in the quadratic polynomial, we would have used ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$.