Question

How do you factor the trinomial `y^2-7y-30`?

Answer 1

`(y+3)(y-10)`

Explanation:

`y^2-7y-30 = y^2-10y+3y-30 = y(y-10)+3(y-10)`

`= (y-10)(y+3) = (y+3)(y-10)`[Ans]

Answer 2

`(y+3)(y-10)`

Explanation:

For a simpler trinomial like this quadratic, in the form `y^2 + ay + b`, one can just look for a pair of numbers which multiply to get `b` and sum to `a`. In this case, they must multiply to get `-30` and add to `-7`, so the numbers are `3` and `-10` because `3*-10 = -30` and `3-10 = -7`.
If you can't immediately see that then you can use another method like completing the square:
`y^2 - 7y -30 = 0`
`(y - \frac{7}{2})^2 - (\frac{7}{2})^2 -30 = 0`
`(y - \frac{7}{2})^2 = \frac{169}{4}`
`y-\frac{7}{2} = +-sqrt\frac{169}{4} = +-\frac{13}{2}`
`y = \frac{7 +- 13}{2}`
`y = \frac{20}{2} or \frac{-6}{2} = 10 or -3`
So if our roots are 10 and -3, the factorisation of it is `(y-10)(y-(-3)) = (y-10)(y+3)`.
I hope I've explained that well enough.

Answer 3

`(y-10)(y+3)`

Explanation:

To factor this, you have to find 2 numbers such that `a+b=-7` and `ab=-30` The 2 numbers, `a` and `b` are `-10` and `3`.

Since `y^2` can be expressed as `(y)(y)`, we get the following :

`y^2-7y-30=(y-10)(y+3)`

Using the FOIL method, we can expand this to check

`y^2+3y-10y-30`