How do you factor completely: $2x^2-10x+12$ ?

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Answer 1 · 2015-07-28T06:30:45

$color(blue)((2)(x-2)(x-3)$ is the factorised form.

$2x^2−10x+12$

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like: $ax^2 + bx + c$ , we need to think of 2 numbers such that:

$N_1*N_2 = a*c = 2*12 =24$
and
$N_1 +N_2 = b = -10$

After trying out a few numbers we get $N_1 = -6$ and $N_2 =-4$
$(-6)*(-4) = 24$ , and $-6+(-4)=-10$

$2x^2−10x+12 = 2x^2−6x -4x+12$

$=2x(x-3) -4(x-3)$

Here, $(x-3)$ is common to both terms:

$=(2x-4)(x-3)$
Also, $2$ is common to both terms of the first bracket

$color(blue)(2(x-2)(x-3)$ is the factorised form.

How do you factor completely: 2x^2-10x+122x^2-10x+12?