NECO 2023 · Paper 2 · Q11

(a)
Solve the equation $\dfrac{1}{x + 1} + \dfrac{2}{x - 1} = \dfrac{1}{x + 3}$ (2 d.p.).
Separate values with commas, e.g. 3, −2
(b)
Find the quotient and remainder when $2x^4 - 9x^3 - 21x^2 + 88x + 48$ is divided by $x - 2$ .
Show the answer
Quotient $2x^3 - 5x^2 - 31x + 26$ , remainder 100
(c)
Given that $p(x) = x^5 + 5x^4 + 9x^3 + 11x^2 + 12x + 13$ , find $3p(2)$ .

Worked solution (try it first)

Put the left side over one denominator: $\dfrac{(x - 1) + 2(x + 1)}{(x + 1)(x - 1)} = \dfrac{3x + 1}{x^2 - 1}$ .
So $\dfrac{3x + 1}{x^2 - 1} = \dfrac{1}{x + 3}$ .
Cross-multiply: $(3x + 1)(x + 3) = x^2 - 1$ .
Expand the left side: $3x^2 + 10x + 3 = x^2 - 1$ .
Take $x^2 - 1$ from both sides: $2x^2 + 10x + 4 = 0$ .
Divide by 2: $x^2 + 5x + 2 = 0$ .
Use the formula: $x = \dfrac{-5 \pm \sqrt{25 - 8}}{2}$
$= \dfrac{-5 \pm \sqrt{17}}{2}$ .
With $\sqrt{17} \approx 4.1231$ : $x \approx -0.44$ or $x \approx -4.56$ .
Neither is $-1$ , $1$ or $-3$ , so both are allowed.

Divide by $x - 2$ with synthetic division: write 2 on the left and the coefficients $2, -9, -21, 88, 48$ .
Bring down 2.
Then $2 \times 2 = 4$ and $-9 + 4 = -5$ .
$-5 \times 2 = -10$ and $-21 - 10 = -31$ .
Next $-31 \times 2 = -62$ and $88 - 62 = 26$ .
Then $26 \times 2 = 52$ and $48 + 52 = 100$ .
So the quotient is $2x^3 - 5x^2 - 31x + 26$ and the remainder is 100.
Check: $f(2) = 32 - 72 - 84 + 176 + 48 = 100$ ✓.

Substitute $x = 2$ term by term: $p(2) = 32 + 5(16) + 9(8) + 11(4) + 12(2) + 13$ .
So $p(2) = 32 + 80 + 72 + 44 + 24 + 13 = 265$ .
Then $3p(2) = 3 \times 265 = 795$ .