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

1 Answer
Sep 2, 2016

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)