Polynomial problem (URGENT)?

Question

Given the polynomial is exactly divisible by x+1, and when it is divided by 3x-1, the remainder is 4. The polynomial gives a remainder hx+k when divided by 3x^2+2x-1.
Find h and k.
(workings needed)

Sorry that is not the correct answer according to my answer given (i need workings)

Answer 1

Well, I worked all morning on this one and finally
came up with an answer for you. It will turn out that
h = 0 and k will depend on a parameter, which
I will explain below. Since you give only
2 conditions on the polynomial, it is impossible to determine
k uniquely.
Let's note several facts.
1). 3x²+2x-1 = (x+1)(3x-1).
(2). Since the polynomial(let's call it P(x) ) gives
a linear remainder when divided by a quadratic,
it is a cubic polynomial.
3). From the conditions of the problem
P(x) = (3x-1)(x²+px+q) + 4.
We also require P(-1) = 0
This gives -4(1-p+q) + 4 = 0.
-p+q = 0.
q = p.
So P(x) = (3x-1)(x² + px + p) +4
= 3x³ + (3p-1)x² + 2px + 4-p
Now divide this synthetically by x + 1 (or use long
division). The quotient is 3x² + (3p-4)x + p+4.
Finally,
P(x)/ (x+1)(3x-1) =
(3x² + (3p-4)x + p+4) )/ (3x -1)
and, by long division, this gives
quotient: x + p-1
remainder: 2p+3.
So the final answer is h = 0, k = 2p+3.
I checked this out for several values of p
and it worked fine. Try it for p = 1.
You get a quotient of x and a remainder of 5.
Hope that helps.

Answer 2

I think that's funny. (x+1)(3x-1) = 3x^2+2x-1. If the polynomial gives a remainder of hx+k when divided by 3x^2+2x-1, and if the polynomial is completely divisible by (x+1), then the remainder is due to (3x-1). This means that (hx+k) = 4. 4 can be written as 0k+4.Thus, h = 0 and k = 4.

Hope that helps.

Answer 3

enable 2 zeroes are a and b a+b =-a million ab = -2 (a-b)^2 = (a+b)^2 -4ab = a million +8=9 a-b = (+/-)3 a+b = -a million a-b =3 a =a million and b =-2 subsequently required poly universal is (x-a million)(x+2){x -3+sqrt(3)}{ x- 3- sqrt 3}=0 (x^2 + x - 2){(x-3)^2 -3}=0 (x^2 +x -2){x^2-6x +9 -3} =0 (x^2 +x-2)(x^2 -6x +6} =0 {x^4 -6x^3 +6x^2}+ {x^3 -6x^2+6x} +{-2x^2+12x-12} =0 x^4 -5x^3 -2x^2+18x-12 = 0 ................Ans