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

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

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Answer 1 · 2016-04-22T16:50:06

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

Explanation:

$4 c^{2} - 12 c + 9$ $4c^2 - 12c + 9$

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

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

$N_{1} \cdot N_{2} = x \cdot z = 4 \cdot 9 = 36$ $N_1*N_2 = x*z = 4 * 9 = 36$

AND

$N_{1} + N_{2} = y = - 12$ $N_1 +N_2 = y = -12$

After trying out a few numbers we get $N_{1} = - 6$ $N_1 = -6$ and $N_{2} = - 6$ $N_2 =-6$
$- 6 \cdot - 6 = 36$ $-6*-6 = 36$ , and $(- 6) + (- 6) = - 12$ $(-6)+(-6)= -12$

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

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

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

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

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

1 Answer

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