WAEC 2018 · Paper 2 · Q11

(a)
(i) Write down the binomial expansion of $\left(2 - \frac12x\right)^5$ 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 of $-17$ when divided by $(x + 2)$ . Find the values of $q$ and $r$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Each term is $\binom5r 2^{5 - r}\left(-\frac12x\right)^r$ for $r = 0, 1, \ldots, 5$ .
$r = 0$ : $32$ .
$r = 1$ : $5 \times 16 \times \left(-\frac12x\right) = -40x$ .
$r = 2$ : $10 \times 8 \times \frac14x^2 = 20x^2$ .
$r = 3$ : $10 \times 4 \times \left(-\frac18x^3\right) = -5x^3$ .
$r = 4$ : $5 \times 2 \times \frac{1}{16}x^4 = \frac58x^4$ .
$r = 5$ : $-\frac{1}{32}x^5$ .
So $\left(2 - \frac12x\right)^5 = 32 - 40x + 20x^2 - 5x^3 + \frac58x^4 - \frac{1}{32}x^5$ .

Find $x$ : $2 - \frac12x = 1.99$ gives $\frac12x = 0.01$ , so $x = 0.02$ .
Substitute: $32 - 40(0.02) + 20(0.02)^2 - 5(0.02)^3 + \ldots = 32 - 0.8 + 0.008 - 0.00004 + \ldots$
That is $31.20796\ldots$ , so $(1.99)^5 = 31.21$ to 2 decimal places.

$(x + 1)$ is a factor, so $f(-1) = 0$ : $-1 + q - r + 9 = 0$ , which gives $q - r = -8$ .
The remainder on dividing by $(x + 2)$ is $-17$ , so $f(-2) = -17$ : $-8 + 4q - 2r + 9 = -17$ , which gives $2q - r = -9$ .
Take the first equation from the second: $q = -1$ .
Then $r = q + 8 = 7$ .