1. ## [SOLVED] Algebraic Division

Normal algebraic division using remainder or factor theorem is usually fine however this exam question asks to find values in a different way:

The polynomial
$f(x) = px^3 - x^2 + qx - 6$
has $(x - 3)$ as a factor. When $f(x)$ is divided by $(x - 2)$ the remainder is -20.
Show that p = 2 and find the value of q.

I can find these values easily by division however the question is worth 6 marks and requires a different approach.
How would you work this question out with that in mind?

Thanks if you can help.

2. Find f(3) and f(2) in terms of p and q. Then solve that resulting system of equations.

3. Originally Posted by db5vry
Normal algebraic division using remainder or factor theorem is usually fine however this exam question asks to find values in a different way:

The polynomial
$f(x) = px^3 - x^2 + qx - 6$
has $(x - 3)$ as a factor. When $f(x)$ is divided by $(x - 2)$ the remainder is -20.
Show that p = 2 and find the value of q.

I can find these values easily by division however the question is worth 6 marks and requires a different approach.
How would you work this question out with that in mind?

Thanks if you can help.
If $x - 3$ is a factor, then the remainder when $f(x)$ is divided by $x - 3$ is $0$. Therefore $f(3) = 0$.

If $f(x)$ is divided by $x - 2$, the remainder is $-20$. So $f(2) = -20$.

Using these pieces of information...

$f(3) = p(3)^3 - 3^2 + q(3) - 6 = 0$

$27p - 9 + 3q - 6 = 0$

$27p + 3q = 15$

and

$f(2) = p(2)^3 - 2^2 + q(2) - 6 = -20$

$8p - 4 + 2q - 6 = -20$

$8p + 2q = -10$.

So now you have 2 equations in 2 unknowns which you can solve simultaneously.

$27p + 3q = 15$
$8p + 2q = -10$

$54p + 6q = 30$
$24p + 6q = -30$

So $30p = 60$

$p = 2$.

If $8p + 2q = -10$

$8(2) + 2q = -10$

$16 + 2q = -10$

$2q = -26$

$q = -13$.

4. Originally Posted by Prove It
If $x - 3$ is a factor, then the remainder when $f(x)$ is divided by $x - 3$ is $0$. Therefore $f(3) = 0$.

If $f(x)$ is divided by $x - 2$, the remainder is $-20$. So $f(2) = -20$.

Using these pieces of information...

$f(3) = p(3)^3 - 3^2 + q(3) - 6 = 0$

$27p - 9 + 3q - 6 = 0$

$27p + 3q = 15$

and

$f(2) = p(2)^3 - 2^2 + q(2) - 6 = -20$

$8p - 4 + 2q - 6 = -20$

$8p + 2q = -10$.

So now you have 2 equations in 2 unknowns which you can solve simultaneously.

$27p + 3q = 15$
$8p + 2q = -10$

$54p + 6q = 30$
$24p + 6q = -30$

So $30p = 60$

$p = 2$.

If $8p + 2q = -10$

$8(2) + 2q = -10$

$16 + 2q = -10$

$2q = -26$

$q = -13$.
Thank you for explaining that, it was really clear and it is very greatly appreciated!