# help please - polynomial factors - how to solve (not the answer)

• Nov 7th 2012, 01:27 PM
indymogul
help please - polynomial factors - how to solve (not the answer)
Hi, I'm just looking for a suggestion on how to go about solving this. I'm not looking for the answer to the question, I have that in the back of the textbook. This is in the review of the chapter Algebra 1. I've covered all the sections and done all the exercises with only a few questions I couldn't solve on my own. I just really don't know where to go with this one. The question is:

If x2 + bx -2 is a factor of x3 + (2b - 1)x2 - p, find the two possible values of p (where p is a real number).

I'm not looking for the answer to the question. I'd really appreciate a hint on how to tackle the problem myself. I've covered linear simultaneous equations, polynomials, factors inc, the factor theorem, factorising cubics and rational functions so far. I feel like I should be able to do this but I'm really stuck with how to proceed with it. I tried long division but was unable to get past the first line. Do I need to simplify first? Is it possible to solve it this way?

Thanks.
• Nov 7th 2012, 01:44 PM
HallsofIvy
Re: help please - polynomial factors - how to solve (not the answer)
Because $x^2+ bx- 2$ is of degree 2, and $x^3+ (2b-1)x^2- p$ is of degree 3, if the first is a factor of the second, then it must be of the form
$(x^2+ bx- 2)(x- c)= x^3+ (2b-1)x^2- p$. Multiplying out the left side, $x^3+ (b- c)x^2- (2+ bc)x+ 2c= x^3+ (2b-1)x^2- p$. Since that is to be true for all x, we must have b- c= 2b- 1, 2+ bc= 0, and 2c= -p. Solve the last equation for c= -p/2 and replace c with that in the previous two equations. You should be able to reduce those two equations to a single quadratic equation for p.
• Nov 10th 2012, 11:48 AM
indymogul
Re: help please - polynomial factors - how to solve (not the answer)
Thanks very much for this. Using your advice I was able to solve for one value of b. Starting from (b - c) = (2b -1), I got b - c + 1 = 2b, and then b = 1 - c. From (2 + bc) = 0, I arrived at c = (-2/b). Combining these 2, I got b^2 = b + 2. I just had to look at this to realize that b = 2. But was unable to get the other value for b, which I later worked out was -1. The book gives the 2 possible answers for p, 2 and -4. I was able to solve for p being 2, but not for -4. When I went back I realized that the other possible value for b was -1, because if b^2 = b + 2, then b can equal -1 because -1^2 = -1 + 2. But I was unable to solve for this value, and technically I also did not solve for the first value of b (2) because I just looked at the equation b^2 = b + 2 and knew it was 2.

My question is, how could I have correctly solved for both values? I tried several possible rearrangements of the various equations resulting from (b - c) = (2b -1), (2 + bc) = 0 but I was unable to arrive at a definitive value for either variable. Thanks, and sorry for my ineptitude, I'm out of school a long time.
• Nov 10th 2012, 12:35 PM
skeeter
Re: help please - polynomial factors - how to solve (not the answer)
Quote:

My question is, how could I have correctly solved for both values? I tried several possible rearrangements of the various equations resulting from (b - c) = (2b -1), (2 + bc) = 0 but I was unable to arrive at a definitive value for either variable.
$b-c = 2b-1 \implies b = 1-c$

sub $(1-c)$ into the second equation ...

$2+(1-c)c = 0 \implies c^2-c-2 = 0 \implies (c-2)(c+1) = 0$

two possible values for $c$ ... since $2c= -p$ , two possible values for $p$
• Nov 10th 2012, 03:00 PM
indymogul
Re: help please - polynomial factors - how to solve (not the answer)
perfect. thank you skeeter and halls of ivy! it's starting to make sense to me now. i'm halfway through the revision of algebra 1 now and no further problems.