Division of polinomials.

thraiye · **Joined:** Sun, 25 Dec 2011 08:28:42 UTC **Posts:** 12

Find the values of "a" and "b" if the
expression (x*x*x)+a(x*x)+bx-4 is exactly divisible by (x*x)-4.

Thanks for any possible solution because dividing such equations is extremely confusing to me.

Shadow · **Posted:** Wed, 27 Jun 2012 13:45:09 UTC

thraiye wrote:

Find the values of "a" and "b" if the
expression (x*x*x)+a(x*x)+bx-4 is exactly divisible by (x*x)-4.

Thanks for any possible solution because dividing such equations is extremely confusing to me.

What equations? Neither of those has an equals sign...

Check out http://www.sosmath.com/algebra/factor/fac01/fac01.html for a nice primer, and also http://www.ebsinstitute.com/Baseball/EBS.crm5df2.html .

Soroban · **Posted:** Wed, 27 Jun 2012 15:23:44 UTC

Hello, thraiye!

Quote:

$\text{Find the values of }a\text{ and }b\text{ if the expression }P(x) \:=\:x^3+ax^2+bx-4$
. . $\text{is divisible by }x^2-4$

If

divides $P(x),\,\text{ then }x = \pm2\text{ are zeros of }P(x).$

$\text{We have: }\:\begin{array}{ccccccc}x = 2: & 8 + 4a + 2b + 4 &=& 0 & [1] \\ x = \text{-}2: & \text{-}8 + 4a - 2b + 4 &=& 0 & [2] \end{array}$

$\begin{array}{cccccccccc}\text{Add [1] and [2]:} & 8a + 8 \:=\: 0 & \Rightarrow & \boxed{a \:=\: \text{-}1} \\ \\[-4mm] \text{Subtract [1] - [2]:} & 16 + 4b \:=\: 0 & \Rightarrow & \boxed{b \:=\: \text{-}4} \end{array}$

outermeasure · **Posted:** Wed, 27 Jun 2012 15:30:05 UTC

Soroban wrote:

Hello, thraiye!

Quote:

$\text{Find the values of }a\text{ and }b\text{ if the expression }P(x) \:=\:x^3+ax^2+bx-4$
$\text{is divisible by }x^2-4$

If

divides $P(x),\,\text{ then }x = \pm2\text{ are zeros of }P(x).$

$\text{We have: }\:\begin{array}{ccccccc}x = 2: & 8 + 4a + 2b + 4 &=& 0 & [1] \\ x = \text{-}2: & \text{-}8 + 4a - 2b + 4 &=& 0 & [2] \end{array}$

$\begin{array}{cccccccccc}\text{Add [1] and [2]:} & 8a + 8 \:=\: 0 & \Rightarrow & \boxed{a \:=\: \text{-}1} \\ \\[-4mm] \text{Subtract [1] - [2]:} & 16 + 4b \:=\: 0 & \Rightarrow & \boxed{b \:=\: \text{-}4} \end{array}$

The constant term is -4, not +4.

Denis · **Posted:** Wed, 27 Jun 2012 16:49:02 UTC

If x=5, a=3, b=7 (as example):

X^3 + ax^2 + bx - 4 = 231
x^2 - 4 = 21
231 / 21 = 11 remainder 0

Soooo....what am I missing?

Soroban · **Posted:** Wed, 27 Jun 2012 18:44:49 UTC

For some reason, I'm not allowed to edit my post.
So, I'll correct my error in this new post.

Quote:

$\text{Find the values of }a\text{ and }b\text{ if the expression }P(x) \:=\:x^3+ax^2+bx-4$
. . $\text{is divisible by }x^2-4$

If

divides $P(x),\,\text{ then }x = \pm2\text{ are zeros of }P(x).$

$\text{We have: }\:\begin{array}{ccccccc}x = 2: & 8 + 4a + 2b - 4 &=& 0 & [1] \\ x = \text{-}2: & \text{-}8 + 4a - 2b - 4 &=& 0 & [2] \end{array}$

$\begin{array}{cccccccccc}\text{Add [1] and [2]:} & 8a - 8 \:=\: 0 & \Rightarrow & \boxed{a \:=\: 1} \\ \\[-4mm] \text{Subtract [1] - [2]:} & 16 + 4b \:=\: 0 & \Rightarrow & \boxed{b \:=\: \text{-}4} \end{array}$

thraiye · **Joined:** Sun, 25 Dec 2011 08:28:42 UTC **Posts:** 12

Thanks a lot everyone.

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