In this question you need to use the information given to form and then solve two linear simultaneous equations

$p (x)$ has a factor $(x - 2)$ , so from the factor theorem we know $x = 2$ is a solution of $p (x) = 0$ .
Substitute $x = 2$ into $p (x) = 0$

$6 {(2)}^{3} + a {(2)}^{2} + b (2) + 2 = 0$

Now simplify to get an equation in terms of $a$ and $b$ .

$6 \times 8 + 4 a + 2 b + 2 = 0 48 + 4 a + 2 b + 2 = 0$

[1]

We can also collect like terms and pass the constant term to the right hand side

$4 a + 2 b = - 50$

As all the terms are even, we can also divide both sides by 2

$2 a + b = - 25 1$

We now have one of two simultaneous equations

To get the other, use "p(1) = -2p(0)". Substitute 1 into p to get p(1), substitute 0 into p to get p(0)

$\begin{array}{rcl} p (1) & = & 6 {(1)}^{3} + a {(1)}^{2} + b (1) + 2 \\ = & 6 + a + b \\ p (0) & = & 6 {(0)}^{3} + a {(0)}^{2} + b (0) + 2 \\ = & 2 \end{array}$

p(1) = -2p(0) so equate p(1) to negative two times p(0)

$\begin{array}{rcl} 6 + a + b & = & 2 \times 2 \end{array}$

[1]

Simplify this and pass the constants to the right hand side to get the second of the two simultaneous equations

$\begin{array}{rcl} 6 + a + b & = & 4 \\ a + b & = & - 2 2 \end{array}$

Solve the simultaneous equations by elimination as below or by substitution; whichever method you feel most comfortable with!

$\begin{array}{rcl} 2 a + b & = & - 25 1 \\ a + b & = & - 12 2 \end{array}$

Subtract $2$ from $1$ , taking care with the negative numbers

$\begin{array}{rcl} 1 - 2 : a & = & - 13 \end{array}$

any valid attempt to solve simultaneous equations [1]

And substitute $a = - 13$ into $1$ or $2$ ( $2$ is easier) to obtain $a$

$\begin{array}{rcl} a + b & = & - 12 \\ - 13 + b & = & - 12 \\ b & = & 1 \end{array}$

Make sure that both the values of $a$ and $b$ are clearly written.

$a = - 13, b = 1$ [1]

Factors of Polynomials (Cambridge O Level Additional Maths)

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