How do you factor $b^{2} + 4 b - 60$ $b^2+4b-60$ ?

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Answer 1 · 2015-05-02T05:08:24

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:

$N_{1} \cdot N_{2} = a \cdot c = 1 \cdot - 60 = - 60$ $N_1*N_2 = a*c = 1*-60 = -60$
AND
$N_{1} + N_{2} = b = 4$ $N_1 +N_2 = b = 4$

After trying out a few numbers we get $N_{1} = 10$ $N_1 = 10$ and $N_{2} = - 6$ $N_2 = - 6$

$10 \cdot - 6 = 60$ $10*-6 = 60$ , and $10 + (- 6) = 4$ $10+(-6) = 4$

$b^{2} + 4 b - 60 = b^{2} + 10 b - 6 b - 60$ $b^2+4b-60 = b^2 + 10b - 6b - 60$

$= b (b + 10) - 6 (b + 10)$ $= b(b+10) - 6(b+10)$

$b + 10$ $b + 10$ is a common factor to each of the terms

$= (b + 10) (b - 6)$ $=color(green)( (b+10)(b-6)$

How do you factor b^2+4b-60b2+4b−60b^2+4b-60?