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

3 Answers
Jun 6, 2017

(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]

Jun 6, 2017

(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.

Jun 6, 2017

(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