(a) Differentiate ((x – 3)(x^{2} + 5)) with respect to x.
(b) If ((x + 1)^{2}) is a factor of (f(x) = x^{3} + ax^{2} + bx + 3), where a and b are constants, find the :
(i) values of a and b ; (ii) zeros of f(x).
Explanation
(a) Let (u = (x – 3)) and (v = (x^{2} + 5)).
(y = uv)
(frac{mathrm d u}{mathrm d x} = 1 ; frac{mathrm d v}{mathrm d x} = 2x)
From the product rule,
(frac{mathrm d y}{mathrm d x} = v frac{mathrm d u}{mathrm d x} + u frac{mathrm d v}{mathrm d x})
(frac{mathrm d y}{mathrm d x} = (x^{2} + 5)(1) + (x – 3)(2x))
= (x^{2} + 5 + 2x^{2} – 6x)
= (3x^{2} – 6x + 5)
(b)(i) (f(x) = x^{3} + ax^{2} + bx + 3)
Since ((x + 1)^{2}) is a factor, then (f(-1) = 0).
(therefore f(-1) = (-1)^{3} + a(-1)^{2} + b(-1) + 3)
(0 = -1 + a – b + 3 implies a – b = -2 … (1))
When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. Since we already have 2 out of the three zeros of the polynomial, the third one will be (x + 3). [Check for yourself using the long division method].
(therefore f(-3) = 0)
(f(-3) = (-3)^{3} + a(-3)^{2} + b(-3) + 3)
(-27 + 9a – 3b + 3 = 0 implies 9a – 3b = 24 … (2))
From (1), (a = -2 + b)
(therefore 9(-2 + b) – 3b = 24)
(-18 + 9b – 3b = 24 implies 6b = 42)
(b = 7 implies a = -2 + 7 = 5)
(a, b) = (5, 7).
(ii) We already have the three factors of the polynomial as ((x + 1)^{2}; (x + 3)).
Hence the zeros of f(x) are -1, -1 and -3.