How do you factor completely $6c^2-33c+15$ ?

Question

1445 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-29T15:01:19

$(6c - 3) ( c - 5)$ is the factorised form of the expression.

$6c^2 - 33c + 15$

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

In this technique, if we have to factorise an expression like $xc^2 + yc + z$ , we need to think of 2 numbers such that:

$N_1*N_2 = x * z = 6 * 15 = 90$

AND

$N_1 +N_2 = y = -33$

After trying out a few numbers we get $N_1 = -30$ and $N_2 =-3$
$(-30 )* (-3) = 90$ , and $(-30 )+( - 3)= -33$

$6c^2 - 33c + 15 = 6c^2 - 30c - 3c + 15$

$= 6c ( c - 5) - 3 ( c +5)$

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

How do you factor completely 6c^2-33c+156c^2-33c+15?