(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).
(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.
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