Is it possible to factor y=x^2-4x-3 ? If so, what are the factors?

Feb 16, 2017

Factors are $y = \left(x - 2 + \sqrt{7}\right) \left(x - 2 - \sqrt{7}\right)$

Explanation:

When you generally say - Ïs it possible to factor $y = a {x}^{2} + b x + c$, what is normally meant is to have rational factors of type $y = a \left(x + p\right) \left(x + q\right)$

This is possible if the discriminant ${b}^{2} - 4 a c$ is square of a rational number.

Here in $y = {x}^{2} - 4 x - 3$, we have $a = 1$, $b = - 4$ and $c = - 3$ and hence ${b}^{2} - 4 a c = {\left(- 4\right)}^{2} - 4 \times 1 \times \left(- 3\right) = 16 + 12 = 28$

as it is not the square of a rational number you cannot have rational factors. But it is still possible to have irrational factors.

As $y = {x}^{2} - 4 x - 3 = \left({x}^{2} - 2 \times 2 \times x + {2}^{2}\right) - {2}^{2} - 3$

= ${\left(x - 2\right)}^{2} - 7 = {\left(x - 2\right)}^{2} - {\left(\sqrt{7}\right)}^{2}$

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

factors are $y = \left(x - 2 + \sqrt{7}\right) \left(x - 2 - \sqrt{7}\right)$.