Question

How do you factor `-6a^2-25a-25`?

Answer 1

factor y = -6a^2 - 25a - 25

Explanation:

`y = -(6a^2 + 25a + 25)` = a(x - p)(x - q)
I use the new AC method.
Converted `y' = a^2 + 25 a + 150. = (x - p')(x - q')`
Factor pairs of 150 -> (5, 30)(10, 15). This sum is 25 = b.
Then, p' = 10 and q' = 15
Then, `p = (p')/a = 10/6 = 5/3` and `q = (q')/a = 15/6 = 5/2`

`y = -6(x + 5/3)(x + 5/2) = - (3x + 5)(2x + 5)`

Answer 2

`-6a^2-25a-25 = (-1)(2a+5)(3a+5)`

Explanation:

[check out Nghi N's answer; this version is only an alternate explanation... perhaps using what Nghi would call the "old AC method", but without the `p` and `q` (plain and primed) variables.]

Before getting into the details, factor out the obvious `(-1)`
and replace the variable `a` with `x` (since the "ac method" uses `a` for something else
`color(white)("XXXX")` `-6x^2-25x-25 = (-1)(6x^2+25x+25)`
and we will focus on factoring only this second term.

The "ac methd" says that given a polynomial of the form
`color(white)("XXXX")` `ax^2+bx+c`
then the polynomial can be factored if there exists a pair of factors (m,n) of `ac` such that:
`color(white)("XXXX")` `{ (m+n = b if c>0), (m-n = b if c<0) :}`

`ac = 6xx25 = 150`
`150` can be factored into primes as `2xx3xx5xx5`
from which we can derive the factor pairs
`(2xx75), (3xx50), (5xx30), (6xx25), (10xx15)`

Since `c=25>0` we are looking for a factor pair whose sum is `25 (=b)`
and the only pair that meets this condition is
`color(white)("XXXX")` `(10,15)`

Rewrite `6x^2+25x+25` splitting the middle term into two terms each with a coefficient of one of the factors:
`color(white)("XXXX")` `6x^2+10x+15x+25`

Factor by grouping:
`color(white)("XXXX")` `(6x^2+10x)+(15x+25)`
`color(white)("XXXX")` `2x(3x+5)+5(3x+5)`
`color(white)("XXXX")` `(2x+5)(3x+5)`

Restoring the `(-1)` and `a` in place of `x`
`color(white)("XXXX")` `(-1)(2a+5)(3a+5)`