How do you factor the expression $3 x^{2} + 4 x - 4$ $3x^2+4x-4$ ?

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Answer 1 · 2016-03-01T18:32:37

$(3 x - 2) (x + 2)$ $(3x-2)(x+2)$ is the factorised form of the equation.

$3 x^{2} + 4 x - 4$ $3x^2 +4x-4$

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

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

$N_{1} \cdot N_{2} = a \cdot c = 3 \cdot - 4 = - 12$ $N_1*N_2 = a*c = 3*-4 = -12$
AND
$N_{1} + N_{2} = b = 4$ $N_1 +N_2 = b = 4$

After trying out a few numbers we get $N_{1} = 6$ $N_1 = 6$ and $N_{2} = - 2$ $N_2 =-2$

$6 \cdot - 2 = - 12$ $6*-2 = -12$ , and $6 + (- 2) = 4$ $6+(-2)= 4$

$3 x^{2} + 4 x - 4 = 3 x^{2} + 6 x - 2 x - 4$ $3x^2 +4x-4 =3x^2 +6x-2x-4$

$= 3 x (x + 2) - 2 (x + 2)$ $= 3x(x+2) - 2(x+2)$

$= (3 x - 2) (x + 2)$ $= (3x-2)(x+2)$

$(3 x - 2) (x + 2)$ $(3x-2)(x+2)$ is the factorised form of the equation.

How do you factor the expression 3x2+4x−43x^2+4x-4?