# Math Help - Dividing Polynomials: Remainder.

1. ## 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:

2. 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 $a=f_2(k)$.

3. Hello, anshulbshah!

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

And emakarov is correct!

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

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

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

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

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:

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

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

5. 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.