[SOLVED] Algebraic Division

• Mar 8th 2009, 03:56 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
\$\displaystyle f(x) = px^3 - x^2 + qx - 6\$
has \$\displaystyle (x - 3)\$ as a factor. When \$\displaystyle f(x)\$ is divided by \$\displaystyle (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, 03:59 PM
wytiaz
Find f(3) and f(2) in terms of p and q. Then solve that resulting system of equations.
• Mar 8th 2009, 04:07 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
\$\displaystyle f(x) = px^3 - x^2 + qx - 6\$
has \$\displaystyle (x - 3)\$ as a factor. When \$\displaystyle f(x)\$ is divided by \$\displaystyle (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 \$\displaystyle x - 3\$ is a factor, then the remainder when \$\displaystyle f(x)\$ is divided by \$\displaystyle x - 3\$ is \$\displaystyle 0\$. Therefore \$\displaystyle f(3) = 0\$.

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

Using these pieces of information...

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

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

\$\displaystyle 27p + 3q = 15\$

and

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

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

\$\displaystyle 8p + 2q = -10\$.

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

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

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

So \$\displaystyle 30p = 60\$

\$\displaystyle p = 2\$.

If \$\displaystyle 8p + 2q = -10\$

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

\$\displaystyle 16 + 2q = -10\$

\$\displaystyle 2q = -26\$

\$\displaystyle q = -13\$.
• Mar 8th 2009, 04:19 PM
db5vry
Quote:

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

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

Using these pieces of information...

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

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

\$\displaystyle 27p + 3q = 15\$

and

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

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

\$\displaystyle 8p + 2q = -10\$.

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

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

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

So \$\displaystyle 30p = 60\$

\$\displaystyle p = 2\$.

If \$\displaystyle 8p + 2q = -10\$

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

\$\displaystyle 16 + 2q = -10\$

\$\displaystyle 2q = -26\$

\$\displaystyle q = -13\$.

Thank you for explaining that, it was really clear and it is very greatly appreciated! (Hi)