How do you factor 3y2+21y−36?

2 Answers
Alan P.
May 16, 2015

3y2+21y−36

First extract the (obvious) constant factor of 3
3(y2+7y−12)

Unfortunately (y2+7y−12) has no obvious rational roots.

We can apply the quadratic formula for roots:
x=−b±√b2−4ac2a
to get
x=−7±√972

So with irrational factors we could continue to
3(x+7+√972)(x+7−√972))

George C.
May 16, 2015

First notice that all the coefficients are divisible by 3

3y2+21y−36=3(y2+7y−12)

To factor y2+7y−12, notice that it is of the form ay2+by+c, with a=1, b=7 and c=−12.

By the general quadratic solution, this is zero for

y=−b±√b2−4ac2a

=−7+=√72−4⋅1⋅(−12)2⋅1

=−7±√49+482

=−7±√972

97 is not a perfect square, so the only factorization into linear factors has irrational coefficients, viz.

y2+7y−12

=(y−(−7+√972))(y−(−7−√972))

So in summary:

3y2+21y−36

=3(y−(−7+√972))(y−(−7−√972))

The factorization would have been much simpler if we were finding the factors of 3y2+21y+36 or 3y2−21y+36.

In those cases, we have:

3y2+21y+36=3(y+3)(y+4)

3y2−21y+36=3(y−3)(y−4)

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