How do you factor $14 y^{2} + 15 y - 9$ $14y^2+15y-9$ ?

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How do you factor $14 y^{2} + 15 y - 9$ $14y^2+15y-9$ ?

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Answer 1 · 2015-08-11T16:38:59

$(7 y - 3) (2 y + 3)$ $color(blue)(( 7y-3) (2y+3)$ is the factorised form of the expression.

Explanation:

$14 y^{2} + 15 y - 9$ $14y^2+15y−9$

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

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

$N_{1} \cdot N_{2} = a \cdot c = 14 \cdot - 9 = - 126$ $N_1*N_2 = a*c = 14*-9 = -126$
and,
$N_{1} + N_{2} = b = 15$ $N_1 +N_2 = b = 15$

After trying out a few numbers we get $N_{1} = 21$ $N_1 = 21$ and $N_{2} = - 6$ $N_2 =-6$
$21 \cdot (- 6) = - 126$ $21*(-6) = -126$ , and $21 + (- 6) = 15$ $21+(-6)= 15$

$14 y^{2} + 15 y - 9 = 14 y^{2} + 21 y - 6 y - 9$ $14y^2+color(blue)(15y)−9 = 14y^2+color(blue)(21y -6y)−9$

$= 7 y (2 y + 3) - 3 (2 y + 3)$ $= 7y(2y+3) -3 (2y +3)$

$(7 y - 3) (2 y + 3)$ $color(blue)(( 7y-3) (2y+3)$ is the factorised form of the expression.

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How do you factor $14 y^{2} + 15 y - 9$ $14y^2+15y-9$ ?

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