Question

How do you factor `7y^2+19y+10`?

Answer 1

`7y^2+19y+10 = (7y+5)(y+2)`

Explanation:

If `7y^2+19y+10` can be factored into two binomials:
`color(white)("XXXX")` `(ay+b)(cy+d)`
`color(white)("XXXX")` `color(white)("XXXX")` `=acy^2+(ad+bc)y+bd`
then we need to find factors
`color(white)("XXXX")` `a, c` of `7`
and
`color(white)("XXXX")` `b, d` of `10`
such that
`color(white)("XXXX")` `ad+bc=19`

The only integer factors of `7` are
`color(white)("XXXX")` `(a,c) = (7,1)`
The (non-negative) integer factors of `10` are
`color(white)("XXXX")` `(b,d) in {(1,10), (10,1), (2,5),(5,2)}`

This gives us 4 possible combinations to check and we find:
`color(white)("XXXX")` `ad + bc = (7xx2) + (5xx1)` gives the required `19`

So `(ay+b)(cy+d)` becomes `(7y+5)(1y+2)`

Answer 2

Factor: `y = 7y^2 + 19y + 10`

Ans: (7x + 5)(x + 2)

Explanation:

I use the new AC Method.
Converted `y' = y^2 + 19y + 70 =`(y - p')(y - q').
p' and q' have same sign (Rule of signs)
Factor pairs of 70 --> (2, 35)(5, 14). This sum is 19 = b.
Then, p' = 5 and q' = 14.
Therefor, `p = (p')/a = 5/7,` and `q = 14/7 = 2`

Factored form: y = 7(y + 5/7)(y + 2) = (7y + 5)(y + 2).