Question

How do you factor the trinomial `10y^2+22y-24`?

Answer 1

`2(5y-4)(y+3)`

Explanation:

First, recognize that there is a common factor of `2` in each term, which can be factored out of the expression.

`10y^2+22y-24=2(5y^2+11y-12)`

To factor the remaining trinomial `5y^2+11y-12`, we need to look for two numbers that meet the following criteria:

• Their sum is the coefficient of the middle term `mathbf((11))`
• Their product is the product of the coefficients of the first and last terms `mathbf("(")5xx-12=mathbf(-60")")`

We should examine the factor pairs of `-60` and try to find a pair whose sum is `11`.

`{:("-1 and 60",",","1 and -60"),("-2 and 30",",","2 and -30"),("-3 and 20",",","3 and -20"),(color(blue)"-4 and 15",",","4 and -15"),("-5 and 12",",","5 and -12"),("-6 and 10",",","6 and -10"):}`

We should use the pair `-4,15` to rewrite `11y`:

`2(5y^2+11y-12)=2(5y^2+15y-4y-12)`

Now, factor by grouping: factor a common terms from each smaller "set" of two in the term:

`2(color(red)(5y^2+15y)-color(green)(4y-12))=2[color(red)((5y^2+15y))-color(green)((4y+12))]`

Notice that the sign of the `12` changed since a negative was factored from the second term.

`2[(5y^2+15y)-(4y+12)]=2[5y(y+3)-4(y+3)]`

Factor a common `(y+3)` term from the `5y` and `-4` terms. Recall the `2` will stay "outside," just hanging around.

`2[5y(y+3)-4(y+3)]=ul(2(5y-4)(y+3)`

This is fully factored.

Answer 2

`f(y) = 2(5y - 4)(y + 3)`

Explanation:

`f(y) = 2f'(y) =2(5y^2 + 11y - 12)`
I use the new AC Method to factor trinomials (Socratic Search)
`f'(y) = 5y^2 + 11y - 12 =` 5(x + p)(x + q)
Converted trinomial `f''(y) = y^2 + 11y - 60 =` (y + p')(y + q').
p' and q' have opposite signs.
Compose factor pairs of (ac = -60) --> ...(-3, 20)(-4, 15). This sum is 11 = b. Then, p' = -4 and q' = 15.
Back to original trinomial f'(y): p = (p')/a = -4/5 and q = (q')/a = 15/5 = 3.
Factored form: `f(y) = 2(5)(y - 4/5)(y + 3) =` 2(5y - 4)(y + 3)