How do you factor #4y=x^3-4x^2-11x+30 #?

1 Answer
Sep 2, 2016

Answer:

#4y = (x-2)(x-5)(x+3)#

Explanation:

By the rational roots theorem, any rational zeros of #x^3-4x^2-11x+30# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #30# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-5, +-6, +-10, +-15, +-30#

If #x=2# then we find:

#x^3-4x^2-11x+30 = (8)-4(4)-11(2)+30 = 8-16-22+30 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#x^3-4x^2-11x+30 = (x-2)(x^2-2x-15)#

To factor the remaining quadratic, find a pair of numbers with difference #2# and product #15#. The pair #5, 3# works, hence:

#x^2-2x-15 = (x-5)(x+3)#

Putting it all together, we have:

#4y = (x-2)(x-5)(x+3)#