How do you factor completely $2x^2+20=x^2+9x$ ?

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Answer 1 · 2017-12-28T20:03:45

GIven: $2x^2+20=x^2+9x$

Combine terms so that the quadratic is equal to 0:

$x^2-9x+20=0$

This will factor into $(x-r_1)(x-r_2) = 0$ , if we can find numbers such that $r_1r_2 = 20$ and $-(r_1+r_2) = -9$ .

4 and 5 will do it $(4)(5) = 20$ and $-(4+5) = -9$ :

$(x - 4)(x-5) = 0$

Answer 2 · 2017-12-28T20:08:15

Set it equal to zero, then find factors of $c$ that add to $b$ .

Set equal to zero:

$x^2-9x+20=0$

Looking at the discriminate: ( $b^2-4ac$ )

$81-4*1*20=81-80=1$

Since $1$ is a perfect square, we know it factors.

Factors of $20$ that add to $-9$ are $-5$ and $-4$ .

$(x-5)(x-4)=x^2-9x+20$

How do you factor completely 2x^2+20=x^2+9x2x^2+20=x^2+9x?