How do you factor completely: $3x^2- 21x + 30$ ?

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Answer 1 · 2015-07-24T13:38:02

$color(blue)((3x-6)(x-5)$ is the factorised form.

$3x^2-21x+30$

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:

$N_1*N_2 = a*c = 3*30 =90$
and
$N_1 +N_2 = b = -21$

After trying out a few numbers we get:
$N_1 = -15$ and $N_2 =-6$
$(-15)*(-6) = 90$ , and $-15+(-6)= -21$

$3x^2-21x+30 =3x^2-15x - 6x+30$

$=3x(x-5) - 6(x-5)$

$(x-5)$ is common to both terms
$color(blue)((3x-6)(x-5)$ is the factorised form.

How do you factor completely: 3x^2- 21x + 303x^2- 21x + 30?