Real zeros and polynomial #15: P(x) = x^3 + 3x^2 - 4

[Illvoices]

Find all rational zeros of the polynomial, and write the polynomial in factored form.
15. P(x)=x³+3x²-4

so could someone tell me how'd the answer for this question is -2,1
when i got 1,-1 and where did the -2 come from where you are suppose to find the leading coefficient and the constant term which is 1 plus or minus and 4 plus or minus.

 

[ksdhart2]

...you are suppose to find the leading coefficient and the constant term which is 1 plus or minus and 4 plus or minus.

It seems to me like you're trying to apply the rational root theorem, but it's all getting a bit jumbled in your head, because what I've quoted above is close to what the theorem says, but not quite. Accordingly, a good next step might be to return to the definition and review exactly what it says. For a cubic polynomial of the form Ax³ + Bx² + Cx + D, the rational root theorem tells us that any rational roots of the polynomial will be in this form:

\(\displaystyle \pm \dfrac{\text{Factors of D}}{\text{Factors of A}}\)

For your specific polynomial, you have D = -4 and A = 1. That actually makes things a lot easier because the only factors of A are +/- 1, so any rational roots discovered by this theorem will simply be the factors of D, which are 1, -1, 2, -2, 4, and -4. Now, I trust you can finish up from here.

 

[Illvoices]

thank you ive got the answer i wanted, although if you could tell me what does p and q stand for in the zero theorem?

 

[ksdhart2]

Well, the written version of the theorem takes many forms, and I suspect every textbook is slightly unique in their terminology. The version I learned used neither the letter p, nor q. However, my best guess would be you saw it in the context of saying that any rational root of f(x) will be of the form \(\displaystyle \dfrac{p}{q}\). If that's the case, then p would simply be a stand in for "any factor of D" and q a stand in for "any factor of A." If the variables you're talking about are used in a different context, then I haven't a clue.

 

[Illvoices]

what i thought i did was solve with -1 in parentheses:
P(-1)=(-1)³+3(-1)²-4=
p(-1)=-1+-3=-4-4=0

 

[Otis]

P(-1)=(-1)³+3(-1)²-4

p(-1)=-1+-3=-4-4=0

3 times (-1)^2 is not -3

Also, don't switch back and forth between names P and p. Use one or the other, not both.

Also also, you don't need to type exercise numbers in the body of your posts. (If you're trying to keep track, just put it in the subject line.) In another thread, some people misread your exercise number as part of the problem.

 

[Illvoices]

OK thanks peeps ive finished with the problem i'll get going with my next problem now.