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

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

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Answer 1 · 2017-12-28T20:03:45

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

Combine terms so that the quadratic is equal to 0:

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

This will factor into $(x - r_{1}) (x - r_{2}) = 0$ $(x-r_1)(x-r_2) = 0$ , if we can find numbers such that $r_{1} r_{2} = 20$ $r_1r_2 = 20$ and $- (r_{1} + r_{2}) = - 9$ $-(r_1+r_2) = -9$ .

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

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

Answer 2 · 2017-12-28T20:08:15

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

Explanation:

Set equal to zero:

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

Looking at the discriminate: ( $b^{2} - 4 a c$ $b^2-4ac$ )

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

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

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

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

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

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