How do you factor the quadratic expression completely? $2 x^{2} - 13 x + 20$ $2x^2 - 13x + 20$

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How do you factor the quadratic expression completely? $2 x^{2} - 13 x + 20$ $2x^2 - 13x + 20$

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Answer 1 · 2018-06-10T16:35:42

$(x - 4) (2 x - 5)$ $(x-4)(2x-5)$

Explanation:

$2 x^{2} - 13 x + 20$ $2x^{2}-13x+20$

Factors of $20 \cdot 2$ $20 * 2$ which add up to -13 are -5 and -8

so you can replace -13 with -5 and -8 such that:

$2 x^{2} - 5 x - 8 x + 20$ $2x^{2}-5x-8x+20$
which goes to:
$x (2 x - 5) - 4 (2 x - 5)$ $x(2x-5)-4(2x-5)$

Take the expressions not in the brackets together to get:
$(x - 4) (2 x - 5)$ $(x-4)(2x-5)$

Answer 2 · 2018-06-10T16:36:29

$(x - 4) (x - \frac{5}{2})$ $(x-4)(x-5/2)$

Explanation:

$2 x^{2} - 13 x + 20$ $2x^2-13x+20$

The simplest way to do this is using the quadratic equation:

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

Where,
$a = 2$ $a=2$
$b = - 13$ $b=-13$
$c = 20$ $c=20$

$x = \frac{- (- 13) \pm \sqrt{{(- 13)}^{2} - 4 \cdot 2 \cdot 20}}{2 \cdot 2}$ $x=(-(-13)+-sqrt((-13)^2-4*2*20))/(2*2)$

$x = \frac{(13) \pm \sqrt{169 - 160}}{4}$ $x=((13)+-sqrt(169-160))/(4)$
$x = \frac{(13) \pm \sqrt{9}}{4}$ $x=((13)+-sqrt(9))/(4)$
$x = 4$ $x=4$ and $x = \frac{5}{2}$ $x=5/2$

$x - 4 = 0$ $x-4=0$
$x - \frac{5}{2} = 0$ $x-5/2=0$

Therefore,

$2 x^{2} - 13 x + 20 = (x - 4) (x - \frac{5}{2})$ $2x^2-13x+20= (x-4)(x-5/2)$

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How do you factor the quadratic expression completely? $2 x^{2} - 13 x + 20$ $2x^2 - 13x + 20$

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