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Thread: Dividing Polynomials: Remainder.

  1. #1
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    Question 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:
    please help at the earliest...have exams day after tom...
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  2. #2
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    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 $\displaystyle a=f_2(k)$.
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  3. #3
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    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]: .$\displaystyle a \:=\:5^2 - 8(5) + 10 \quad\Rightarrow\quad \boxed{a \:=\:-5}$

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  4. #4
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    Thanks .. Got my mistake
    Quote Originally Posted by Soroban View Post
    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]: .$\displaystyle a \:=\:5^2 - 8(5) + 10 \quad\Rightarrow\quad \boxed{a \:=\:-5}$

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  5. #5
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    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.
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