# [SOLVED] Algebraic Division

• Mar 8th 2009, 04:56 PM
db5vry
[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.
• Mar 8th 2009, 04:59 PM
wytiaz
Find f(3) and f(2) in terms of p and q. Then solve that resulting system of equations.
• Mar 8th 2009, 05:07 PM
Prove It
Quote:

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 $8p + 2q = -10$

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

$16 + 2q = -10$

$2q = -26$

$q = -13$.
• Mar 8th 2009, 05:19 PM
db5vry
Quote:

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! (Hi)