(a)(i) Write down the binomial expansion of ((2 – frac{1}{2}x)^{5}) in ascending powers of x.
(ii) Using the expansion in (a)(i), find, correct to two decimal places, the value of ((1.99)^{5}).
(b) The polynomial (x^{3} + qx^{2} + rx + 9), where q and r are constants, has (x + 1) as a factor and has a remainder -17 when divided by (x + 2). Find the values of q and r.
Explanation
(a)(i) ((2 – frac{1}{2}x)^{5} = ^{5}C_{5}(2)^{5}(-frac{1}{2}x)^{0} + ^{5}C_{4}(2)^{4}(-frac{1}{2}x)^{1} + ^{5}C_{3}(2)^{3}(-frac{1}{2}x)^{2} + ^{5}C_{2}(2)^{2}(-frac{1}{2}x)^{3} + ^{5}C_{1} (2)^{1}(-frac{1}{2}x)^{4} + ^{5}C_{0} (2)^{0}(-frac{1}{2}x)^{5})
= (32 – 5(8x) + 10(2x^{2}) – 10(frac{x^{3}}{2}) + 5(frac{x^{4}}{8}) – frac{x^{5}}{32})
= (32 – 40x + 20x^{2} – 5x^{3} + frac{5}{8}x^{4} – frac{1}{32}x^{5})
(ii) ((1.99)^{5} = (2 – frac{1}{2}x)^{5})
(2 – frac{1}{2}x = 1.99 implies frac{1}{2}x = 2 – 1.99 = 0.01)
(x = 0.01 times 2 = 0.02)
Put x = 0.02 in the equation, we have
(32 – 40(0.02) + 20(0.02)^{2} – 5(0.02)^{3} + frac{5}{8}(0.02)^{4} – frac{1}{32}(0.02)^{5})
(32 – 0.8 + 0.008 – …)
= (31.208 approxeq 31.21)
(b) (f(x) = x^{3} + qx^{2} + rx + 9)
(x + 1) is a factor, hence, f(-1) = 0.
(f(-1) = (-1)^{3} + q(-1)^{2} + r(-1) + 9)
(0 = -1 + q – r + 9)
(-8 = q – r … (1))
(x + 2) has remainder -17, therefore f(-2) = -17.
(-17 = (-2)^{3} + q(-2)^{2} + r(-2) + 9)
(-17 = -8 + 4q – 2r + 9)
(-18 = 4q – 2r implies -9 = 2q – r … (2))
From (1), r = q + 8.
(therefore 2q – q – 8 = -9)
(q = -1)
(r = -1 + 8 = 7)
((q, r) = (-1, 7))