(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 (2frac{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} = 2frac{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})