(a) The polynomial \(f(x) = x^{3} + px^{2} - 10x +... | WAEC SSCE - Further Mathematics - 2012 (Essay)

Question 1 Report

(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.

Answer

(a) Since \(x^2 + x - 6 = (x+3)(x-2)\), the polynomial \(f(x) = x^3 + px^2 - 10x + q\) is zero at \(x = -3\) and \(x = 2\).

(i) Using \(f(2) = 0\): \(8 + 4p - 20 + q = 0 \Rightarrow 4p + q = 12\).

Using \(f(-3) = 0\): \(-27 + 9p + 30 + q = 0 \Rightarrow 9p + q = -3\).

Subtracting: \(5p = -15 \Rightarrow p = -3\), and \(q = 12 - 4(-3) = 24\).

So \(p = -3,\; q = 24\).

(ii) With \(f(x) = x^3 - 3x^2 - 10x + 24\), divide by \(x^2 + x - 6\):

\[x^3 - 3x^2 - 10x + 24 = (x^2 + x - 6)(x - 4).\]

The third factor is \((x - 4)\).

(b) For a cube of side \(s\), volume \(V = s^3\), so \(\dfrac{dV}{dt} = 3s^2\dfrac{ds}{dt}\).

Given \(\dfrac{dV}{dt} = 2\tfrac12 = 2.5\ \text{cm}^3\text{s}^{-1}\) and \(s = 2\ \text{cm}\):

\[2.5 = 3(2)^2\frac{ds}{dt} = 12\frac{ds}{dt} \Rightarrow \frac{ds}{dt} = \frac{2.5}{12} = \frac{5}{24} \approx 0.21\ \text{cm s}^{-1}.\]

The side is increasing at \(\dfrac{5}{24}\ \text{cm s}^{-1}\).

Read lesson note on Polynomial Functions (WAEC)

Answer Details