(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 de...
Question 1 Report
(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.
Answer Details
(a)(i) Expand \(\left(2-\tfrac12 x\right)^{5}\) using \(\binom{5}{k}2^{5-k}\left(-\tfrac12 x\right)^{k}\):
\[32-40x+20x^{2}-5x^{3}+\tfrac58 x^{4}-\tfrac{1}{32}x^{5}\]
(ii) Put \(2-\tfrac12 x=1.99\Rightarrow \tfrac12 x=0.01\Rightarrow x=0.02\). Substitute:
\(32-40(0.02)+20(0.02)^{2}-5(0.02)^{3}+\cdots\)
\(=32-0.8+0.008-0.00004+\cdots\approx31.20796\).
\[(1.99)^{5}\approx31.21\ \text{(2 d.p.)}\]
(b) Let \(P(x)=x^{3}+qx^{2}+rx+9\).
\((x+1)\) is a factor \(\Rightarrow P(-1)=0\): \(-1+q-r+9=0\Rightarrow q-r=-8\).
Remainder \(-17\) on division by \((x+2)\Rightarrow P(-2)=-17\): \(-8+4q-2r+9=-17\Rightarrow 4q-2r=-18\Rightarrow 2q-r=-9\).
Subtracting: \((2q-r)-(q-r)=-9-(-8)\Rightarrow q=-1\), then \(r=q+8=7\).
\[\boxed{q=-1,\ r=7}\]
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