## [SPLIT] solving y(y+1)(y+2)(y+3) = 7920

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### [SPLIT] solving y(y+1)(y+2)(y+3) = 7920

I'm trying to solve y(y+1)(y+2)(y+3) = 7920, which is a problem from my friend's kid. First I multiplied it all out:

(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

I tried to solve it as a quadratic equation, but that didn't seem to work since this polynomial has 5 terms. I think it is 3 terms to be a trinomial.

Since this equation can't be grouped as a trinomial, I am not able to solve it by factoring. Am I on the right track? I haven't done any maths for more than 3 years, and I think I forgot almost everything I learned!

Can you please give me a hint?
mathlearner2011

Posts: 1
Joined: Thu Jun 23, 2011 3:08 am

### Re: [SPLIT] solving y(y+1)(y+2)(y+3) = 7920

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
nona.m.nona

Posts: 250
Joined: Sun Dec 14, 2008 11:07 pm

### Re: [SPLIT] solving y(y+1)(y+2)(y+3) = 7920

$7920 = 2^4\cdot 3^2\cdot 5\cdot 11=8\cdot9\cdot10\cdot11$

therefore $y=8$

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: [SPLIT] solving y(y+1)(y+2)(y+3) = 7920

Martingale wrote:$7920 = 2^4\cdot 3^2\cdot 5\cdot 11=8\cdot9\cdot10\cdot11$

therefore $y=8$

also note ...

$7920=8\cdot9\cdot10\cdot11=(-8)\cdot(-9)\cdot(-10)\cdot(-11)=(-11)\cdot(-10)\cdot(-9)\cdot(-8)$

so $y$ could also be -11

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA