How do you factor completely $4c^2 - 12c + 9$ ?

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Answer 1 · 2016-04-22T16:50:06

$(2c - 3 ) (2c -3)$ is the factorised form of the expression.

$4c^2 - 12c + 9$

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:

$N_1*N_2 = x*z = 4 * 9 = 36$

AND

$N_1 +N_2 = y = -12$

After trying out a few numbers we get $N_1 = -6$ and $N_2 =-6$
$-6*-6 = 36$ , and $(-6)+(-6)= -12$

$4c^2 - 12c + 9 = 4c^2 - 6c - 6c + 9$

$= 2c(2c -3) -3 (2c - 3)$

$(2c - 3 )$ is a common factor to each of the terms

$= (2c - 3 ) (2c -3)$

How do you factor completely 4c^2 - 12c + 94c^2 - 12c + 9?