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

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

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Answer 1 · 2015-05-28T15:34:07

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

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like $a z^{2} + b z + c$ $az^2 + bz + c$ , we need to think of 2 numbers such that:
$N_{1} \cdot N_{2} = a \cdot c = 9 \cdot 4 = 36$ $N_1*N_2 = a*c = 9*4 = 36$
and
$N_{1} + N_{2} = b = - 12$ $N_1 +N_2 = b = -12$
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 $(- 6) + (- 6) = - 12$ $(-6)+(-6)= -12$

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

$= 3 z (3 z - 2) - 2 (3 z - 2)$ $=3z(3z - 2) -2(3z-2)$
$= (3 z - 2) (3 z - 2)$ $=color(green)((3z-2)(3z-2)$

note : We know that ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $color(green)((a - b)^2 = a^2- 2ab + b^2)$

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

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