Question 4
Problem:
$f(x) = x^3 + px^2 + qx + 6$ where $p$ and $q$ are constants.
Given that $(x - 1)$ is a factor of $f(x)$ and that when $f(x)$ is divided by $(x + 1)$, the remainder is 8.
(a)
(i) Show that $p = -2$
(ii) Find the value of $q$
(6 marks)
(b)
Hence, solve the equation $f(x) = 0$
(3 marks)
Problem Solution
We are given the polynomial:
$$ f(x) = x^3 + px^2 + qx + 6 $$
where $p$ and $q$ are constants. It is given that $(x - 1)$ is a factor of $f(x)$, and when $f(x)$ is divided by $(x + 1)$, the remainder is 8.
Part (a) (i): Show that $p = -2$
Since $(x - 1)$ is a factor of $f(x)$, by the Factor Theorem, we know:
$$ f(1) = 0 $$
Substitute $x = 1$ into $f(x)$:
$$ f(1) = 1^3 + p(1)^2 + q(1) + 6 = 1 + p + q + 6 = 0 $$
This simplifies to:
$$ p + q + 7 = 0 \quad \Rightarrow \quad p + q = -7 \quad \text{(Equation 1)} $$
Part (a) (ii): Find the value of $q$
Next, we are told that when $f(x)$ is divided by $(x + 1)$, the remainder is 8. By the Remainder Theorem, we know:
$$ f(-1) = 8 $$
Substitute $x = -1$ into $f(x)$:
$$ f(-1) = (-1)^3 + p(-1)^2 + q(-1) + 6 = -1 + p - q + 6 = 8 $$
Simplifying:
$$ p - q + 5 = 8 \quad \Rightarrow \quad p - q = 3 \quad \text{(Equation 2)} $$
Solving for $p$ and $q$
We now solve the system of equations:
$$ p + q = -7 \quad \text{(Equation 1)} $$
$$ p - q = 3 \quad \text{(Equation 2)} $$
Add these two equations:
$$ (p + q) + (p - q) = -7 + 3 $$
$$ 2p = -4 \quad \Rightarrow \quad p = -2 $$
Substitute $p = -2$ into Equation 1:
$$ -2 + q = -7 \quad \Rightarrow \quad q = -5 $$
Part (b): Solve the equation $f(x) = 0$
Now that we know $p = -2$ and $q = -5$, the polynomial becomes:
$$ f(x) = x^3 - 2x^2 - 5x + 6 $$
Since $(x - 1)$ is a factor, divide $f(x)$ by $(x - 1)$ using synthetic division:
$$ \begin{array}{r|rrrr} 1 & 1 & -2 & -5 & 6 \\ & & 1 & -1 & -6 \\ \hline & 1 & -1 & -6 & 0 \\ \end{array} $$
The quotient is $x^2 - x - 6$, and the remainder is 0, confirming that $x - 1$ is a factor. Thus:
$$ f(x) = (x - 1)(x^2 - x - 6) $$
Factor $x^2 - x - 6$:
$$ x^2 - x - 6 = (x - 3)(x + 2) $$
Thus:
$$ f(x) = (x - 1)(x - 3)(x + 2) $$
The solutions to $f(x) = 0$ are:
$$ x = 1,\; x = 3,\; x = -2 $$
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