How do you factor the trinomial y^2-7y-30y2−7y−30?

3 Answers
Binayaka C.
Jun 6, 2017

(y+3)(y-10)(y+3)(y−10)

Explanation:

y^2-7y-30 = y^2-10y+3y-30 = y(y-10)+3(y-10) y2−7y−30=y2−10y+3y−30=y(y−10)+3(y−10)

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

Jacob
Jun 6, 2017

(y+3)(y-10)(y+3)(y−10)

Explanation:

For a simpler trinomial like this quadratic, in the form y^2 + ay + by2+ay+b, one can just look for a pair of numbers which multiply to get bb and sum to aa. In this case, they must multiply to get -30−30 and add to -7−7, so the numbers are 33 and -10−10 because 3*-10 = -303⋅−10=−30 and 3-10 = -73−10=−7.
If you can't immediately see that then you can use another method like completing the square:
y^2 - 7y -30 = 0y2−7y−30=0
(y - \frac{7}{2})^2 - (\frac{7}{2})^2 -30 = 0(y−72)2−(72)2−30=0
(y - \frac{7}{2})^2 = \frac{169}{4}(y−72)2=1694
y-\frac{7}{2} = +-sqrt\frac{169}{4} = +-\frac{13}{2}y−72=±√1694=±132
y = \frac{7 +- 13}{2}y=7±132
y = \frac{20}{2} or \frac{-6}{2} = 10 or -3y=202or−62=10or−3
So if our roots are 10 and -3, the factorisation of it is (y-10)(y-(-3)) = (y-10)(y+3)(y−10)(y−(−3))=(y−10)(y+3).
I hope I've explained that well enough.

SoySoy4444
Jun 6, 2017

(y-10)(y+3)(y−10)(y+3)

Explanation:

To factor this, you have to find 2 numbers such that a+b=-7a+b=−7 and ab=-30ab=−30 The 2 numbers, aa and bb are -10−10 and 33.

Since y^2y2 can be expressed as (y)(y)(y)(y), we get the following :

y^2-7y-30=(y-10)(y+3)y2−7y−30=(y−10)(y+3)

Using the FOIL method, we can expand this to check

y^2+3y-10y-30y2+3y−10y−30

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