Dividing Polynomials: Remainder.

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September 18th 2010, 09:07 AM
anshulbshah

Dividing Polynomials: Remainder.

If the $X^4 - 6x^3 + 16x^2 -25x + 10$ is divided by annother polynomial is divided by another polynomial $x^2 - 2x + k$ ,the reaminder comes out to be $x+a$ ,find k and a:
please help at the earliest...have exams day after tom...
September 18th 2010, 10:23 AM
emakarov

Go ahead and divide $x^4 - 6x^3 + 16x^2 -25x + 10$ by $x^2 - 2x + k$ using long division. The remainder is a linear polynomial $f_1(k)x+f_2(k)$ for some expressions $f_1$ , $f_2$ . On the other hand, the remainder is $x+a$ . Therefore, $f_1(k)=1$ , from where you can find $k$ . Then $a=f_2(k)$ .
September 18th 2010, 10:27 AM
Soroban

Hello, anshulbshah!

Your wording is weird . . . I'll tke a guess at what you meant.

And emakarov is correct!

Quote:

$\text{If }\,x^4 - 6x^3 + 16x^2 -25x + 10\,\text{ is divided by }\,x^2 - 2x + k,$

. . $\text{the remainder is: }\;x+a$ .

$\text{Find }\,k\text{ and }\,a.$

Long Division:

$\begin{array}{cccccccccccc} &&&&&& x^2 & - & 4x & + & (8-k) \\ & & --&--&--&--&--&--&--&--&--- \\ x^2\!-\12x\!+\!k & | & x^4 &-& 6x^3 &+& 16x^2 &-& 25x &+& 10 \\ && x^4 &-& 2x^3 &+& kx^2 \\ && --&--&--&--&--- \\ &&& - & 4x^3 &+& (16\!-\!k)x^2 &-& 25x \\ &&& - & 4x^3 &+& 8x^2 &-& 4kx \\ &&& --&--&--&---&--&--- \\ &&&&&& (8\!-\!k)x^2 &-& (25\!-\!4k)x &+& 10 \\ &&&&&& (8\!-\!k)x^2 &-& 2(8\!-\!k)x &+& k(8-k) \\ &&&&&& ---&--&---&--&--- \\ &&&&&&&& (2k\!-\!9)x &+& 10-k(8\!-\!k)\end{array}$

We have: . $(2k-9)x + (k^2 - 8k + 10) \;=\;x + a$

Equate coefficients: . $\begin{Bmatrix}2k-9 &=& 1 & [1] \\ k^2-8k+10 &=& a & [2] \end{Bmatrix}$

From [1]: . $2k-9\:=\:1 \quad\Rightarrow\quad \boxed{k \:=\:5}$

Substitute into [2]: . $a \:=\:5^2 - 8(5) + 10 \quad\Rightarrow\quad \boxed{a \:=\:-5}$
September 18th 2010, 05:24 PM
anshulbshah

Thanks .. Got my mistake

Quote:

Originally Posted by Soroban

Hello, anshulbshah!

Your wording is weird . . . I'll tke a guess at what you meant.

And emakarov is correct!

Long Division:

$\begin{array}{cccccccccccc} &&&&&& x^2 & - & 4x & + & (8-k) \\ & & --&--&--&--&--&--&--&--&--- \\ x^2\!-\12x\!+\!k & | & x^4 &-& 6x^3 &+& 16x^2 &-& 25x &+& 10 \\ && x^4 &-& 2x^3 &+& kx^2 \\ && --&--&--&--&--- \\ &&& - & 4x^3 &+& (16\!-\!k)x^2 &-& 25x \\ &&& - & 4x^3 &+& 8x^2 &-& 4kx \\ &&& --&--&--&---&--&--- \\ &&&&&& (8\!-\!k)x^2 &-& (25\!-\!4k)x &+& 10 \\ &&&&&& (8\!-\!k)x^2 &-& 2(8\!-\!k)x &+& k(8-k) \\ &&&&&& ---&--&---&--&--- \\ &&&&&&&& (2k\!-\!9)x &+& 10-k(8\!-\!k)\end{array}$

We have: . $(2k-9)x + (k^2 - 8k + 10) \;=\;x + a$

Equate coefficients: . $\begin{Bmatrix}2k-9 &=& 1 & [1] \\ k^2-8k+10 &=& a & [2] \end{Bmatrix}$

From [1]: . $2k-9\:=\:1 \quad\Rightarrow\quad \boxed{k \:=\:5}$

Substitute into [2]: . $a \:=\:5^2 - 8(5) + 10 \quad\Rightarrow\quad \boxed{a \:=\:-5}$
September 18th 2010, 05:25 PM
TheCoffeeMachine

A way to avoid long division -- let:

$\displaystyle \frac{x^4 - 6x^3 + 16x^2 -25x + 10}{x^2 - 2x + k} = (bx^2+cx+d)+\frac{x+a}{x^2 - 2x + k}$

$\displaystyle \Rightarrow {x^4 - 6x^3 + 16x^2 -25x + 10} = (x^2 - 2x + k)(bx^2+cx+d)+(x+a)$

$\displaystyle \Rightarrow {x^4 - 6x^3 + 16x^2 -25x + 10} = bx^4+(c-2b)x^3+(d-2c+kb)x^2+(kc-2d+1)x+kc+a$ $\;\; ^{\bigg\star}$

$^* b = 1 \;\; \therefore \; c-2b = -6 \Rightarrow c = -4 \;\; \therefore \; d-2c+kb = 16 \Rightarrow d = 8-k [/Math] [Math]\therefore \; kc-2d+1 = -25 \Rightarrow \boxed{k = 5} \;\; \therefore \; d = 8-k \Rightarrow d = 3 \;\; \therefore \; kc+a = 10 \Rightarrow \boxed{a = -5}.$

$\star$ The key at this step is of course to add the x in the remainder with the other x terms.
$*$ We compare the coefficients, of course.