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

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Answer 1 · 2016-04-06T11:24:15

$x = \frac{3}{2} or \frac{1}{2}$ $x= 3/2 or 1/2$

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

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

Where $a = 4; b = - 8; c = + 3$ $a=4"; "b=-8"; "c=+3$

and $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $" " x=(-b+-sqrt(b^2-4ac))/(2a)$
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$x = \frac{+ 8 \pm \sqrt{{(- 8)}^{2} - 4 (4) (3)}}{2 (4)}$ $x= (+8+-sqrt((-8)^2-4(4)(3)))/( 2(4))$

$x = 1 \pm \frac{\sqrt{16}}{8}$ $x= 1+-sqrt(16 )/8$

$x = \frac{3}{2} or \frac{1}{2}$ $x= 3/2 or 1/2$

Is it possible to factor y=4x2−8x+3y=4x^2 - 8x + 3 ? If so, what are the factors?