Maths Elective IGCSE

f(x) = -x ³ + bx ² + cx + d
f(x) = 0 with x ₀ = 0, x ₁ = 1, x ₂ = k

Because of x ₀ we know: d = 0

So we have the equation f(x) = -x ³ + bx ² + cx
f(1) = -(1 ³) + b(1 ²) + c1 = 0 b + c = 1 b = 1-c

Remainder of division is 8, so we conclude:
(-x ³ + bx ² + cx):(x-3) = -x ² + x(b-3) + (c + 3(b - 3)) + 3(c + 3(b - 3))/(x-3)

3(c + 3(b - 3)) = 8 = 3c + 9b - 27 3c + 9b = 35

Insert b = 1-c into the last equation and solve for c:
3c + 9(1-c) = 35 = 3c + 9 - 9c = 9 - 6c c = -26/6 = -13/3

Insert c = -13/3 into f(1):
b = 1-(-13/3) = 16/3

So we have the final equation: f(x) = -x ³ + (16x ²)/3 - 13x/3 = -x(x ² - 16x/3 + 13/3)

Now just solve x ² - 16x/3 + 13/3 = 0

[input]x^2 - 16x/3 + 13/3 = 0[/input]

Edit: Misread. Actual roots are x ₀ = 1, x ₁ = 2, x ₂ = k
So we have:
f(1) = -(1 ³) + b(1 ²) + c1 + d = 0 b + c + d = 1
f(2) = -(2 ³) + b(2 ²) + 2c + d = -8 + 4b + 2c + d = 0 4b + 2c + d = 8
f(k) = -(k ³) + b(k ²) + kc + d = 0

So the actual division would result into:
(-x ³ + bx ² + cx + d):(x-3) = -x ² + x(b-3) + (c + 3(b - 3)) + (3(c + 3(b - 3)) + d)/(x-3)

(3(c + 3(b - 3)) + d) = 8 9b + 3c + d = 35

Gauss-Jordan:
[1] 1 1 1 1
[2] 4 2 1 8
[3] 9 3 1 35

[2] - [1], [3] - [1]
[1] 1 1 1 1
[2] 3 1 0 7
[3] 8 2 0 34

[3] - 2x[2]
[1] 1 1 1 1
[2] 3 1 0 7
[3] 2 0 0 20 2b = 20 b = 10

Insert b = 10 into [2]
3*10 + c = 7 c = 7-30 = -23

Insert b = 10 and c = -23 into [1]
10 - 23 + d = 1 d = 14

Final equation f(x) = -x ³ + 10x ² - 23x + 14

Since the roots of the polynomial are 1,2, and k, and the lead coefficient is -1 the factors of the polynomial are going to be (-x+1),(x-2), and (x-k).

Therefore f(x)=(-x+1)(x-2)(x-k)

(-x+1)(x-2)(x-k)=-x^3 + (-k+3)x^2 + (3k-2)x + 2k

Dividing this polynomial by (x-3) using synthetic division we get a remainder of 2k-6.

We know that f(x) divided by (x-3) has a remainder of 8. So, 2k-6=8 k=7

If k=7 then f(x)=-x^3+10x^2-23x+14. You can check by dividing this polynomial by (x-3) and seeing if you get a remainder of 8.

Hope that helps.