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When divided by (x-1) the polynomial P(x)=x^3 + ax^2 + bx +12 leaves a remainder of 6. When divided by (x+3), P(x) leaves a remainder of -30.

a) Find the values of the connstants a and b
i got a= -3 and b= -4

b) Show that (x-3) is a factor of P(x)
done that cos P(x)=0

c) Hence find the roots of the equation P(x)=0

i don't get part C........what does it mean by "find the roots" and how do you find the roots???

thanx!
2. find the roots simply means find the solutions
3. but when i find the solution my answer doesn't match the real answer
4. (Original post by Saf!)
but when i find the solution my answer doesn't match the real answer
My solutions are x = 3, 2 or -2
5. a) two equations; solve simultaneously
b) F(3) = 0 , therefore it's a factor
c) Use factor theorem to find the other two roots (or one if it's a repeating root)
6. o.k i feel like a complete dumbo!........but could someone please do part C for me so i can see how it's done?? cos i keep getting the wrong answer
7. x^3 - 3x^2 -4x +12 = (x-3)(x^2 + kx - 4)

lhs = -3x^2
rhs = -3x^2 + kx^2
divide by x [lhs = rhs]
Therefore; -3 = -3 + k
k = 0

Are you sure A & B are right?

Because if they aren't this part will be wrong

(x-3)(x^2 + 0x - 4)
(x-3)((x^2)-4)
(x-3)(x-2)(x+2)
x = 3, 2, -2
8. (Original post by Saf!)
o.k i feel like a complete dumbo!........but could someone please do part C for me so i can see how it's done?? cos i keep getting the wrong answer
P(x)=x^3-3x^2-4x+12 = x^2(x-3) - 4(x-3)
=(x-3)(x^2-4)=(x-3)(x-2)(x+2)
So the roots of P(x) are 3, 2 and -2
9. Thanx guys!.......i started out the right way but got muddled up near the end!
10. No worries, there's like 6 different ways you can do it.

I used the 'k method', make sure you show all working out though!
Long division is probs best method, cause you need to use it for c3 in similiar situations. But that's some boring stuff.

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Updated: February 2, 2005
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