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

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

Postby mathlearner2011 on Thu Jun 23, 2011 3:45 am

Please help me with this. I really confused about the calculation

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

Postby nona.m.nona on Thu Jun 23, 2011 11:21 am

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

Postby Martingale on Thu Jun 23, 2011 12:44 pm



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

Postby Martingale on Thu Jun 23, 2011 6:27 pm

Martingale wrote:

therefore



:) also note ...



so could also be -11
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