mathlearner2011 wrote:I'm trying to solve y(y+1)(y+2)(y+3) = 7920....
(y^2 + y)(y + 2)(y + 3) = 7920
(y^3 + 3y^2 + 2y)(y + 3) = 7920
y^4 + 6y^3 + 11y^2 + 6y - 7920 = 0
This is correct.
mathlearner2011 wrote:I tried to solve it as a quadratic equation....
Quadratics are equations of degree 2, not degree 4.
mathlearner2011 wrote:Since this equation can't be grouped as a trinomial, I am not able to solve it by factoring....
First, apply the Rational Roots Test
to obtain a listing of possible roots (zeroes, solutions). Then use synthetic division
to determine which of the possible roots are actual roots. Once you have found two, you will be down to a quadratic, to which you can apply the Quadratic Formula
For further information on this method, please review the following:Solving Polynomials