How do you factor 3y^2+21y-36?

2 Answers
May 16, 2015

3y^2+21y-36

First extract the (obvious) constant factor of 3
3(y^2+7y-12)

Unfortunately (y^2+7y-12) has no obvious rational roots.

We can apply the quadratic formula for roots:
x=(-b+-sqrt(b^2-4ac))/(2a)
to get
x= (-7 +-sqrt(97))/2

So with irrational factors we could continue to
3(x+(7+sqrt(97))/2)(x+(7-sqrt(97))/2))

May 16, 2015

First notice that all the coefficients are divisible by 3

3y^2 + 21y -36 = 3(y^2 + 7y - 12)

To factor y^2+7y-12, notice that it is of the form ay^2+by + c, with a=1, b=7 and c=-12.

By the general quadratic solution, this is zero for

y = (-b +-sqrt(b^2 - 4ac))/(2a)

= (-7 +=sqrt(7^2-4*1*(-12)))/(2*1)

= (-7 +- sqrt(49+48)) / 2

= (-7 +-sqrt(97))/2

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

y^2+7y-12

= (y-((-7 + sqrt(97))/2))(y-((-7 - sqrt(97))/2))

So in summary:

3y^2 + 21y -36

= 3(y-((-7 + sqrt(97))/2))(y-((-7 - sqrt(97))/2))

The factorization would have been much simpler if we were finding the factors of 3y^2 + 21y + 36 or 3y^2 - 21y + 36.

In those cases, we have:

3y^2 + 21y + 36 = 3(y+3)(y+4)

3y^2 - 21y + 36 = 3(y-3)(y-4)