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

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Answer 1 · 2018-06-10T16:35:42

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

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

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

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

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

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

Answer 2 · 2018-06-10T16:36:29

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

$2x^2-13x+20$

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

$x=(-b+-sqrt(b^2-4ac))/(2a)$

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

$x=(-(-13)+-sqrt((-13)^2-4*2*20))/(2*2)$

$x=((13)+-sqrt(169-160))/(4)$
$x=((13)+-sqrt(9))/(4)$
$x=4$ and $x=5/2$

$x-4=0$
$x-5/2=0$

Therefore,

$2x^2-13x+20= (x-4)(x-5/2)$

How do you factor the quadratic expression completely? 2x^2 - 13x + 202x^2 - 13x + 20