How do you factor 4x^2-5x+6?

Nov 7, 2016

$4 {x}^{2} - 5 x + 6 = \left(2 x - \frac{5}{4} - \frac{\sqrt{71}}{4} i\right) \left(2 x - \frac{5}{4} + \frac{\sqrt{71}}{4} i\right)$

Explanation:

If the sign of the constant term was $-$ rather than $+$ then this would have a factorisation:

$4 {x}^{2} - 5 x - 6 = \left(x - 2\right) \left(4 x + 3\right)$

For the example as given we find:

$4 {x}^{2} - 5 x + 6$

is in the form $a {x}^{2} + b x + c$ with $a = 4$, $b = - 5$ and $c = 6$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 5\right)}^{2} - 4 \left(4\right) \left(6\right) = 25 - 96 = - 71$

Since $\Delta < 0$ this quadratic has no linear factors with Real coefficients.

We can still factor it, but we need Complex coefficients...

$4 {x}^{2} - 5 x + 6 = {\left(2 x\right)}^{2} - 2 \left(\frac{5}{4}\right) \left(2 x\right) + {\left(\frac{5}{4}\right)}^{2} - {\left(\frac{5}{4}\right)}^{2} + 6$

$\textcolor{w h i t e}{4 {x}^{2} - 5 x + 6} = {\left(2 x - \frac{5}{4}\right)}^{2} - \frac{25}{16} + \frac{96}{16}$

$\textcolor{w h i t e}{4 {x}^{2} - 5 x + 6} = {\left(2 x - \frac{5}{4}\right)}^{2} + \frac{71}{16}$

$\textcolor{w h i t e}{4 {x}^{2} - 5 x + 6} = {\left(2 x - \frac{5}{4}\right)}^{2} - {\left(\frac{\sqrt{71}}{4} i\right)}^{2}$

$\textcolor{w h i t e}{4 {x}^{2} - 5 x + 6} = \left(\left(2 x - \frac{5}{4}\right) - \frac{\sqrt{71}}{4} i\right) \left(\left(2 x - \frac{5}{4}\right) + \frac{\sqrt{71}}{4} i\right)$

$\textcolor{w h i t e}{4 {x}^{2} - 5 x + 6} = \left(2 x - \frac{5}{4} - \frac{\sqrt{71}}{4} i\right) \left(2 x - \frac{5}{4} + \frac{\sqrt{71}}{4} i\right)$
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