WAEC 2012 · Paper 2 · Q11

(a)
If $(x + 2)$ and $(x - 1)$ are factors of $f(x) = 6x^4 + mx^3 - 13x^2 + nx + 14$ , find the values of $m$ and $n$ .
Separate values with commas, e.g. 3, −2
(b)
Find the gradient of the circle $x^2 + y^2 - 2x + 2y + 1 = 0$ at the points where $x = 1$ .

Worked solution (try it first)

(a)

$(x + 2)$ is a factor, so $f(-2) = 0$ : $96 - 8m - 52 - 2n + 14 = 0$ , i.e. $4m + n = 29$ .
$(x - 1)$ is a factor, so $f(1) = 0$ : $6 + m - 13 + n + 14 = 0$ , i.e. $m + n = -7$ .
Subtract: $3m = 36$ , so $m = 12$ and $n = -19$ .

(b)

Put $x = 1$ in the circle: $1 + y^2 - 2 + 2y + 1 = 0$ , so $y^2 + 2y = 0$ and $y = 0$ or $y = -2$ .
Differentiate: $2x + 2y\dfrac{dy}{dx} - 2 + 2\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = \dfrac{1 - x}{y + 1}$ .
At $(1, 0)$ and at $(1, -2)$ the top is $1 - 1 = 0$ , so the gradient is $0$ at both points (the tangents are parallel to the $x$ -axis).

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