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June 30, 2015

June 30, 2015

Posted by **mary** on Wednesday, October 29, 2008 at 10:42am.

m^3+5m^2+3m-9

it might be simple i am not seeing it..

please help so i can continue with the problem...thanks

- factoring -
**bobpursley**, Wednesday, October 29, 2008 at 10:47amWell, by inspection, I see that m=1 works, so divide m-1 into the polynomial to see what is left.

- factoring -
**mary**, Wednesday, October 29, 2008 at 11:04amokey i am confused. We are suppose to solve for what m is equal to...i thought we could factor by grouping...but didnt know exactly how..please explain

- factoring -
**bobpursley**, Wednesday, October 29, 2008 at 11:19amone factor is by inspection (m-1)

Divide then m^3+5m^2+3m-9 by (m-1) and you will have the second factor, which I suspect can then be factored easily.

- factoring -
**mary**, Wednesday, October 29, 2008 at 12:00pmi honestly dont see how u divid it and u get the answer...please explain in detail

- factoring -
- factoring -
**Count Iblis**, Wednesday, October 29, 2008 at 1:19pmYou can also use the Rational Roots Theorem. If m = p/q is a solution, then p has to be a divisor of -9 and q a divisor of 1 (-9 is the constant term and 1 is the coefficient of the highest power of m).

This means that all rational solutins must be divisors of 9.

So, you only have to try m = 1, m = -1, m= 3, m = -3, m = 9 and m = -9.

I besides m = 1, m = -3 is a solution. If you had found no other solution besides m = 1, then that would have menat that the other solutioins are not rational numbers. If we find one other, then that means that the remaining one must also be a rational number. But if only m = 1 and m = -3 are the possible rational solutions, this means that one of the roots is a double root.

In this case, you fiund that m = -3 is the double root, and the factorization is:

(m-1)(m+3)^2