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

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Answer 1 · 2015-09-16T15:46:42

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

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

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

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

$N_1*N_2 = a*c = 6*12 = 72$
and
$N_1 +N_2 = b = -17$

After trying out a few numbers we get $N_1 = -9$ and $N_2 =-8$
$(-9)*(-8 )= 72$ , and $-9-8 =-17$

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

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

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

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

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