1. ## Dividing Polynomials: Remainder.

If the $\displaystyle X^4 - 6x^3 + 16x^2 -25x + 10$ is divided by annother polynomial is divided by another polynomial $\displaystyle x^2 - 2x + k$,the reaminder comes out to be $\displaystyle x+a$,find k and a:

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

3. Hello, anshulbshah!

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

And emakarov is correct!

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

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

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

Long Division:

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

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

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

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

Substitute into [2]: .$\displaystyle a \:=\:5^2 - 8(5) + 10 \quad\Rightarrow\quad \boxed{a \:=\:-5}$

4. Thanks .. Got my mistake
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:

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

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

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

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

Substitute into [2]: .$\displaystyle a \:=\:5^2 - 8(5) + 10 \quad\Rightarrow\quad \boxed{a \:=\:-5}$

5. A way to avoid long division -- let:

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

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

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

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

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