(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.
Explanation
(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 \(\frac{5}{2} = 3(2)^{2} \cdot \frac{\mathrm d x}{\mathrm d t}\) \(\frac{\mathrm d x}{\mathrm d t} = \frac{5}{2} \div 12 = \frac{5}{24} cm s^{-1}\)