How do you factor $10 x^{2} + 31 x - 14$ $10x^2+31x-14$ ?

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Answer 1 · 2015-05-06T12:59:53

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:

$N_{1} \cdot N_{2} = a \cdot c = 10 \cdot - 14 = - 140$ $N_1*N_2 = a*c = 10*-14 = -140$
AND
$N_{1} + N_{2} = b = 31$ $N_1 +N_2 = b = 31$

After trying out a few numbers we get $N_{1} = 35$ $N_1 = 35$ and $N_{2} = - 4$ $N_2 =-4$

$35 \cdot - 4 = - 140$ $35*-4 = -140$ , and $35 + (- 4) = 31$ $35+(-4)= 31$

$10 x^{2} + 31 x - 14 = 10 x^{2} + 35 x - 4 x - 14$ $10x^2+31x-14 = 10x^2 +35x - 4x - 14$

$= 5 x (2 x + 7) - 2 (2 x + 7)$ $= 5x(2x+7) - 2(2x+7)$

$2 x + 7$ $2x+7$ is a common factor to each of the terms

$= (2 x + 7) (5 x - 2)$ $=color(green)((2x+7)(5x-2)$

How do you factor 10x^2+31x-1410x2+31x−1410x^2+31x-14?