(i) To find the zeroes of -x^3+9x^2-23x+15, we first apply the factor theorem. Let f(x) = -x^3+9x^2-23x+15.
We check if f(1) = 0:
f(1) = -(1)^3 + 9(1)^2 - 23(1) + 15
f(1) = -1 + 9 - 23 + 15
f(1) = 0
Since f(1) = 0, we know that (x-1) is a factor of f(x). We can then perform polynomial division to find the other factors:
(x-1) * (-x^2 + 10x - 15) = -x^3 + 10x^2 - 15x - x^2 + 10x - 15
= -x^3 + 9x^2 - 5x - 15
From this, we see that -x^2 + 10x - 15 is also a factor. We can then factorize this quadratic to find the remaining zeroes:
-x^2 + 10x - 15 = -(x-3)(x-5)
Therefore, the zeroes of -x^3 + 9x^2 - 23x + 15 are x = 1, 3, and 5.
(ii) To find the zeroes of x^3+2x^2-4x-8, we apply the factor theorem. Let f(x) = x^3+2x^2-4x-8.
We check if f(1) = 0:
f(1) = (1)^3 + 2(1)^2 - 4(1) - 8
f(1) = 1 + 2 - 4 - 8
f(1) = -9
Since f(1) ≠ 0, we know that (x-1) is not a factor of f(x). However, we can still find the zeroes by factoring the polynomial:
x^3 + 2x^2 - 4x - 8 = (x + 2)(x^2 + 1)
Now we solve for the zeroes of x^2 + 1:
x^2 + 1 = 0
x^2 = -1
x = ±√(-1)
x = ±i
Therefore, the zeroes of x^3 + 2x^2 - 4x - 8 are x = -2, i, and -i.
Obtain, using the factor theorem, all the zeroes of :
(i) -x^3+9x^2-23x+15
(ii) x^3+2x^2-4x-8
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