# Is it possible to factor y=x^2+5x-3 ? If so, what are the factors?

Apr 9, 2016

Factors are $\left(x + \frac{5}{2} - \frac{\sqrt{37}}{2}\right) \left(x + \frac{5}{2} + \frac{\sqrt{37}}{2}\right)$

#### Explanation:

In $y = {x}^{2} + 5 x - 3$, the discriminant is given by ${5}^{2} - 4 \cdot 1 \cdot \left(- 3\right) = 25 + 12 = 37$, which is positive but not a square of a rational number.

Hence, we can have irrational factors of $y = {x}^{2} + 5 x - 3$.

As roots of ${x}^{2} + 5 x - 3 = 0$ are $x = \frac{- 5 \pm \sqrt{37}}{2}$

or $x = - \frac{5}{2} + \frac{\sqrt{37}}{2}$ and $x = - \frac{5}{2} - \frac{\sqrt{37}}{2}$

Hence factors of $y = {x}^{2} + 5 x - 3$ are

$\left(x - \left(- \frac{5}{2} + \frac{\sqrt{37}}{2}\right)\right) \left(x - \left(- \frac{5}{2} - \frac{\sqrt{37}}{2}\right)\right)$ or

$\left(x + \frac{5}{2} - \frac{\sqrt{37}}{2}\right) \left(x + \frac{5}{2} + \frac{\sqrt{37}}{2}\right)$