OpenStudy (anonymous):

Given the polynomial f(x)=x^3+ax^2+bx+12 gives a remainder of 32 when divided by (x-5) and a remainder of 20 when divided by (x+1), solve the equation of f(x)=0

OpenStudy (anonymous):

12

OpenStudy (anonymous):

if we put x=0 ..den all d term vanishes and we get f(0)=12

OpenStudy (anonymous):

OpenStudy (anonymous):

i thought what you put at first, but i was mistaken.

OpenStudy (anonymous):

ohh srry ..

OpenStudy (anonymous):

basically you know that f(5) = 32, and f(-1) = 20, you can create a system of equations in a and b. You have 2 equations, 2 unknowns. Then once you figure out a and b, put the values in, and find the zeros of the polynomial by factoring.

OpenStudy (anonymous):

@Joe, yeah ;; I'm at a lost. I think i should do $R_{1}$=32 fx ...I really don't know.

OpenStudy (anonymous):

$f(5) = 32 \iff (5)^3+a(5)^2+b(5)+12 = 32 \iff 25a+5b = -105$

OpenStudy (anonymous):

how do u knw f(5)=32??

OpenStudy (anonymous):

f(-1) = 20 gives a-b = 9

OpenStudy (anonymous):

(x-b) f(b) um.. right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

when you divide f(x) by (x-5) you will get a quotient and a remainder. that looks like this: $\frac{f(x)}{(x-5)} = q(x)+\frac{r(x)}{(x-5)}$ multiplying by x-5 we get: $f(x) = (x-5)q(x)+r(x)$ If you plug in x = 5 you get: $f(5) = (5-5)q(5)+r(5) = r(5) = 32$ since they told us the remainder was 32

OpenStudy (anonymous):

i believe you should get a is -2, and b is -11 from that system. so the polynomial is now: $x^3-2x^2-11x+12$ Now we just need to find the roots.

OpenStudy (anonymous):

1 is a root. by inspection.

OpenStudy (anonymous):

OMG thanks I get it now 25a + 5b = -105 -(25a - 25b= 225) ------------- 30b= -330 b=-330:30 b=-11 is that right...? I'm a bit slow sorry

OpenStudy (anonymous):

im studying for a final atm, so im sry if it seems like im not fullying explaining things. I would be glad to go over again after my test >.< lol

OpenStudy (anonymous):

and yes thats right, b is -11, now you just need a.

OpenStudy (anonymous):

Joe I love you ;w; good luck for your finals!!

OpenStudy (anonymous):

a-(-11)=9 a=9-11 a=-2

OpenStudy (anonymous):

right :) alright im off, good luck with your problem. i'll check it after i get out. thanks for the good luck :)

OpenStudy (anonymous):

You're a genius, thank you~