How do you factor the trinomial $a^{3} - 5 a^{2} - 14 a$ $a^3-5a^2-14a$ ?

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Answer 1 · 2018-03-05T20:05:48

$a (a + 2) (a - 7)$ $a(a+2)(a-7)$

Every term in this trinomial includes an $a$ $a$ , so we can say

$a^{3} - 5 a^{2} - 14 a = a (a^{2} - 5 a - 14)$ $a^3 - 5a^2 - 14a = a(a^2 - 5a - 14)$

All we have to do now is factor the polynomial in brackets, with two numbers that add to $- 5$ $-5$ and multiply to $- 14$ $-14$ .

After some trial and error we find $+ 2$ $+2$ and $- 7$ $-7$ , so

$a^{2} - 5 a - 14 = (a + 2) (a - 7)$ $a^2 - 5a - 14 = (a+2)(a-7)$

so overall we end up with

$a^{3} - 5 a^{2} - 14 a = a (a + 2) (a - 7)$ $a^3 - 5a^2 - 14a = a(a+2)(a-7)$

How do you factor the trinomial a3−5a2−14aa^3-5a^2-14a?