(a) The polynomial \(f(x) = x^{3} + px^{2} - 10x + q\) is exactly divisible by \(x^{2} + x - 6\). Find the :
(i) values of p and q ; (ii) third factor.
(b) The volume of a cube is increasing at the rate of \(2\frac{1}{2} cm^{3} s^{-1}\). Find the rate of change of the side of the base when its length is 2cm.
(a)(i) \(f(x) = x^{3} + px^{2} - 10x + q\)
\(x^{2} + x - 6 = 0 \implies x^{2} - 2x + 3x - 6 = 0\)
\(x(x - 2) + 3(x - 2) = 0 \implies (x - 2)(x + 3) = 0\)
\(x = 2 ; x = -3\)
\(f(-3) = (-3)^{3} + p(-3)^{2} - 10(-3) + q = 0\)
\(-27 + 9p + 30 + q = 0 \implies 9p + q = -3 .... (1)\)
\(f(2) = (2)^{3} + p(2)^{2} - 10(2) + q = 0\)
\(8 + 4p - 20 + q = 0 \implies 4p + q = 12 .... (2)\)
(2) - (1) :
\((4p + q) - (9p + q) = (12 - (-3)) \implies -5p = 15\)
\(p = \frac{15}{-5} = -3\)
Put p = -3 in (2), we have:
\(4(-3) + q = 12 \implies -12 + q = 12 \)
\(q = 12 + 12= 24\)
\(f(x) = x^{3} - 3x^{2} - 10x + 24\)
(ii) Using the method of long division,
\(\frac{x^{3} - 3x^{2} - 10x + 24}{x^{2} + x - 6}\)
we get the third factor = \(x - 4\).
(b) \(\frac{\mathrm d v}{\mathrm d t} = 2\frac{1}{2} cm^{3} s^{-1}\)
\(v = x^{3}\)
\(\frac{\mathrm d v}{\mathrm d x} = 3x^{2}\)
\(\frac{\mathrm d v}{\mathrm d t} = \frac{\mathrm d v}{\mathrm d x} \cdot \frac{\mathrm d x}{\mathrm d t}\)
\(\frac{5}{2} = 3x^{2} \cdot \frac{\mathrm d x}{\mathrm d t}\)
When x = 2cm, we have
\(\frac{5}{2} = 3(2)^{2} \cdot \frac{\mathrm d x}{\mathrm d t}\)
\(\frac{\mathrm d x}{\mathrm d t} = \frac{5}{2} \div 12 = \frac{5}{24} cm s^{-1}\)
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