# Division Algorithm.....

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

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

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...

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

... and that means that is...

$\displaystyle a\ b + d=0$

$\displaystyle c\ d = -c$ (2)

... so that...

$\displaystyle d=-1$

$\displaystyle a\ b=1$ (3)

Kind regards

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

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

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

... and that means that is...

$\displaystyle a\ b + d=0$

$\displaystyle c\ d = -c$ (2)

... so that...

$\displaystyle d=-1$

$\displaystyle a\ b=1$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$