# Division Algorithm.....

• Jun 26th 2010, 04:15 AM
anshulbshah
Division Algorithm.....
If the polynomial f(x) = $ax^3 + bx - c$ is divisible by the polynomial g(x) = $x^2 + bx +c$, then $ab =$ ?
• Jun 26th 2010, 04:41 AM
sa-ri-ga-ma
If f(x) is divisible by g(x), the you can write

$ax^3 + bx - c = (px + q)(x^2 + bx - c)$

Expand the brackets and compare coefficients of like powers of x. Write p and q in terms of a, b and c. Then find ab.
• Jun 26th 2010, 04:45 AM
chisigma
If the hypothesis id true is...

$a\ x^{3} + b\ x -c = (a\ x + d)\ ( x^{2} + b\ x + c)$ (1)

... and that means that is...

$a\ b + d=0$

$c\ d = -c$ (2)

... so that...

$d=-1$

$a\ b=1$ (3)

Kind regards

$\chi$ $\sigma$
• Jun 26th 2010, 05:29 AM
anshulbshah
Quote:

Originally Posted by chisigma
If the hypothesis id true is...

$a\ x^{3} + b\ x -c = (a\ x + d)\ ( x^{2} + b\ x + c)$ (1)

... and that means that is...

$a\ b + d=0$

$c\ d = -c$ (2)

... so that...

$d=-1$

$a\ b=1$ (3)

Kind regards

$\chi$ $\sigma$

@chisigma ....
i ve got 2 qestions regarding this....
1)how did u ax + d to be a multiple of $ax^3 + bx - c$
2) and did u get ab + d by opening the brackets?
• Jun 26th 2010, 06:53 AM
chisigma
Now...

a) if $x^{2} + b\ x + c$ devides $a\ x^{3} + b\ x -c$ , then there is a first order polynomial $a\ x + d$ so that...

$a\ x^{3} + b\ x -c = (a\ x + d)\ (x^{2} + b\ x + c)$ (1)

b) if You develop the product (1) You obtain...

$(a\ x + d)\ (x^{2} + b\ x + c) = a\ x^{3} + (a\ b + d)\ x^{2} + (a\ c + b\ d) x + d\ c = a\ x^{3} + b\ x -c$ (2)

c) from the (2) You obtain ...

$a\ b + d=0$ , $d= -1 \rightarrow a\ b = 1$ (3)

Kind regards

$\chi$ $\sigma$