How do you factor $6 t^{2} - 17 t + 12 = 0$ $6t^2-17t+12=0$ ?

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How do you factor $6 t^{2} - 17 t + 12 = 0$ $6t^2-17t+12=0$ ?

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Answer 1 · 2015-09-16T15:46:42

$(3 t - 4) (2 t - 3)$ $color(blue)((3t-4)(2t - 3)$

Explanation:

$6 t^{2} - 17 t + 12 = 0$ $6t^2 -17t+12=0$

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

In this technique, if we have to factorise an expression like $a t^{2} + b t + c$ $at^2 + bt + c$ , we need to think of 2 numbers such that:

$N_{1} \cdot N_{2} = a \cdot c = 6 \cdot 12 = 72$ $N_1*N_2 = a*c = 6*12 = 72$
and
$N_{1} + N_{2} = b = - 17$ $N_1 +N_2 = b = -17$

After trying out a few numbers we get $N_{1} = - 9$ $N_1 = -9$ and $N_{2} = - 8$ $N_2 =-8$
$(- 9) \cdot (- 8) = 72$ $(-9)*(-8 )= 72$ , and $- 9 - 8 = - 17$ $-9-8 =-17$

$6 t^{2} - 17 t + 12 = 0$ $6t^2 -17t+12=0$

$6 t^{2} - 9 t - 8 t + 12 = 0$ $6t^2 -9t -8t+12=0$

$3 t (2 t - 3) - 4 (2 t - 3) = 0$ $3t(2t - 3) -4(2t-3)=0$

$(3 t - 4) (2 t - 3)$ $color(blue)((3t-4)(2t - 3)$ is the factorised form of the expression.

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How do you factor $6 t^{2} - 17 t + 12 = 0$ $6t^2-17t+12=0$ ?

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How do you factor 6t2−17t+12=06t^2-17t+12=0?

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How do you factor $6 t^{2} - 17 t + 12 = 0$ $6t^2-17t+12=0$ ?