# Is it possible to factor y=4x^2 - 8x + 3 ? If so, what are the factors?

Apr 6, 2016

$x = \frac{3}{2} \mathmr{and} \frac{1}{2}$

#### Explanation:

The coefficient of ${x}^{2}$ is positive so the graph is of format type $\cup$ so the vertex is a minimum.

Standard equation for:$y = a {x}^{2} + b x + c$

Where $a = 4 \text{; "b=-8"; } c = + 3$

and $\text{ } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
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$x = \frac{+ 8 \pm \sqrt{{\left(- 8\right)}^{2} - 4 \left(4\right) \left(3\right)}}{2 \left(4\right)}$

$x = 1 \pm \frac{\sqrt{16}}{8}$

$x = \frac{3}{2} \mathmr{and} \frac{1}{2}$