How do you factor the trinomial $4 a^{2} - a - 5$ $4 a^2-a - 5$ ?

Question

1491 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-17T04:20:15

$(a + 1) (4 a - 5)$ $(a+1)(4a-5)$

This is in standard form, or $a x^{2} + b x + c$ $ax^2 + bx + c$ , where $a = 4$ $a = 4$ , $b = - 1$ $b = -1$ , and $c = - 5$ $c = -5$ .

To factor this, we first have to find $2$ $2$ numbers that:

Add up to $b$ $b$ , or $- 1$ $-1$
Multiply up to $a \cdot c$ $a*c$ , or $4 \cdot - 5$ $4 * -5$ , which equals to $- 20$ $-20$ .

These two numbers are $- 5$ $-5$ and $4$ $4$ .

With these numbers, we can rewrite the trinomial like this:
$4 a^{2} - 5 a + 4 a - 5$ $4a^2 - 5a + 4a - 5$

Now we factor by grouping:
$(4 a^{2} - 5 a) + (4 a - 5)$ $(4a^2 - 5a) + (4a - 5)$

$= a (4 a - 5) + 1 (4 a - 5)$ $=a(4a - 5) + 1(4a-5)$

$= (a + 1) (4 a - 5)$ $=(a+1)(4a-5)$

Hope this helps!

How do you factor the trinomial 4a2−a−54 a^2-a - 5?