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If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.

My Attempt,

$f(x)=x^4-6x^3+16x^2-25x+10$

$g(x)=x^2-2x+k$

$R=x+a$

Here, the divisor is in the quadratic form. so how do I use the synthetic division

pi-π
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  • It is possible to use Synthetic division on a quadratic or higher power with the leading coefficient equal to $1$. In your case, you would ignore the $x^2$ term and write $2,-k$ as the numbers dividing the polynomial – Frank Jan 29 '17 at 02:03
  • @Frank, Could You please show me a bit more? I could not get. – pi-π Jan 29 '17 at 02:07

4 Answers4

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Question: A polynomial $f(x)=x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $g(x)=x^2-2x+k$. Their remainder is $x+a$. Find $k$.

We're given a polynomial, and given that $\frac {f(x)}{g(x)}=p(x)+(x+a)$ where $p(x)$ is a polynomial. Therefore, if we divide $\frac {f(x)}{g(x)}$ and equate the remainder, we should be able to find $k$.

The process for extended synthetic division for higher powers is described below.

\begin{array}{c |c c} & 1 & & -6 & & +16 & & -25 & & +10\\ 2 & & & 2 & & -8 & & 16-2k & & \\-k & & & & & -k & & 4k & & k^2-8k\\\hline\\ &1 & & -4 & & 8-k & & \color{red}{2k-9} & & \color{blue}{k^2-8k+10}\end{array}

Where the top row of numbers are the coefficients of $f(x)$, and the vertical numbers are the negated coefficients of $g(x)$. And the last two numbers being the remainder $\color{red} px+\color{blue} q$.

Comparing the last two numbers to the remainder $x+a$, we have$$(\color{red}{2k-9})x+(\color{blue}{k^2-8k+10})=\color{red}{1}x+\color{blue}{a}\tag1$$$$\implies\begin{cases}2k-9=1\\k^2-8k+10=a\tag2\end{cases}$$ From the first equation of $(2)$, we have$$2k-9=1\implies 2k=10\implies k=5$$And. if necessary, plug in $k=5$ into the second equation of $(2)$ to find $a$.

Frank
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    Re formatting: not exactly my cup of tea, but there are a couple of worked out examples here. – dxiv Jan 29 '17 at 02:25
  • How is the last term in the quotient/eemainder line $0$? Adding up the column I get a quadratic polynomial in $k$, whose value is $-5$ with $k=5$. – Oscar Lanzi Jan 29 '17 at 02:26
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    @OscarLanzi That was a little mistake on my part. It should be fixed. – Frank Jan 29 '17 at 02:27
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    @dxiv I think I have it formatted in $\LaTeX$. :) – Frank Jan 29 '17 at 02:27
  • @Frank, Could you please explain me the process of synthetic division? I could not understand – pi-π Jan 29 '17 at 02:47
  • @NeWtoN Okay. The top row of numbers, the $1,\ -6,\ +16,\ -25,\ +10$ are the coefficients of $f(x)=x^4-6x^3+16x^2-25x+10$. The vertical numbers, the $2,\ -k$ are the negated coefficients of $g(x)=x^2-2x+k$. The polynomial you are dividing into $f(x)$. To actually perform the operation, bring down the $1$ and multiply that by $2$. Place the product on the next column (under $-6$) and subtract. The difference should be $-4$. – Frank Jan 29 '17 at 02:59
  • @Frank, why is there $K$ in the third column? – pi-π Jan 29 '17 at 03:01
  • @NeWtoN Starting with the $-4$, multiply it (again) by $2$ to get $-8$ and place that in the third column under $+16$. However, this time, multiply $-k$ by the previous number before $-4$ ($1$ in this case) and place that in the third column under $+16$. Adding everything, you should get $8-k$. Now, just repeat the process. Extended synthetic division may be difficult, but it's incredibly powerful to use on larger polynomials. It'll definitely take some time to understand the process. It took me about a month to fully understand how to incorporate the extended version of synthetic division. – Frank Jan 29 '17 at 03:01
  • @Frank, Why is $K$ needed to be multiplied in this step? – pi-π Jan 29 '17 at 03:04
  • @NeWtoN What do you mean by "$k$ needed to be multiplied in this step?" – Frank Jan 29 '17 at 03:05
  • @Frank, why was $k$ not kept in 1st and 2nd column? But directly in 3rd column? – pi-π Jan 29 '17 at 03:07
  • @Frank, In the last column, why is $10+(4k-8)+(k^2-8k)=(k^2-8k+10)$?? – pi-π Jan 29 '17 at 03:26
  • @NeWtoN That was a stupid mistake made by me. As for the reason why $k$ does not appear in the first and second column, revisit how Synthetic division is originally derived. Then just extend it to quadratics. – Frank Jan 29 '17 at 03:49
  • @Frank, if the last column is $k^2-4k-8$ then the vvalue of $a\neq -5$, which is the answer in my book? – pi-π Jan 29 '17 at 04:07
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    Looks great, +1. – Oscar Lanzi Jan 29 '17 at 10:49
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You can do the division between $x^4-6x^3+16x^2-25x+10$ and $x^2-2x+k$ following the polynomial long division, getting:

$$R=(2k-9)x+(k^2-8k+10)$$

but $R=x+a$, so $2k-9=1\longrightarrow k=5$ and $k^2-8k+10=a\longrightarrow a=-5$.

MattG88
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HINT

You can write:

$$x^4-6x^3+16x^2-25x+10= (x^2-2x+k)(x^2+bx+c)+x+a=\\ =x^4+(b-2)x^3+(c-2b+k)x^2+(-2c+bk+1)x+(a+kc)$$

So,

$$b-2=-6→b=-4\\ c-2b+k=16\\ -2c+bk+1=-25\\ a+kc=10$$

Arnaldo
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To find the remainder of the long division by $x^2-2x+k$ you can keep replacing $x^2$ with $2x-k$ repeatedly, until getting the remainder of degree $1\,$:

$$ \begin{align} x^4-6x^3+16x^2-25x+10 & = (2x-k)^2 - 6x(2x-k)+16(2x-k)-25x+ 10 \\ & = -8x^2 + (-4k +6k +32-25)x+k^2-16k+10 \\ & = -8(2x-k) + (2k+7)x+k^2-16k+10 \\ & = (2k-9)x + k^2-8k+10 \end{align} $$

Identifying coefficients between the calculated remainder and $x+a$ it follows that $2k-9=1$ so $k=5\,$, and $a=5^2-8 \cdot 5+10=-5\,$.

dxiv
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