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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

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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 – 2017-01-29
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    @Frank, Could You please show me a bit more? I could not get. – 2017-01-29

4 Answers 4

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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 $a$.

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    Re formatting: not exactly my cup of tea, but there are a couple of worked out examples [here](http://meta.math.stackexchange.com/a/21096). – 2017-01-29
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    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 $k=5$. – 2017-01-29
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    @OscarLanzi That was a little mistake on my part. It should be fixed. – 2017-01-29
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    @dxiv I think I have it formatted in $\LaTeX$. :) – 2017-01-29
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    @Frank, Could you please explain me the process of synthetic division? I could not understand – 2017-01-29
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    @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 $-4$. – 2017-01-29
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    @Frank, why is there $K$ in the third column? – 2017-01-29
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    @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. – 2017-01-29
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    @Frank, Why is $K$ needed to be multiplied in this step? – 2017-01-29
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    @NeWtoN What do you mean by "$k$ needed to be multiplied in this step?" – 2017-01-29
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    @Frank, why was $k$ not kept in 1st and 2nd column? But directly in 3rd column? – 2017-01-29
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    @Frank, In the last column, why is $10+(4k-8)+(k^2-8k)=(k^2-8k+10)$?? – 2017-01-29
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    @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. – 2017-01-29
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    @Frank, if the last column is $k^2-4k-8$ then the vvalue of $a\neq -5$, which is the answer in my book? – 2017-01-29
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    Looks great, +1. – 2017-01-29
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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 $k^2-8k+10=a\longrightarrow a=-5$.

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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,

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

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    Where did $b$ come from? – 2017-01-29
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    $b$ and $c$ are coefficient of the quotient. – 2017-01-29
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    I just know that the quotient has degree equal to $2$, so I use a general quadratic polynomial and called its coefficients $b$ and $c$. Is it clear? – 2017-01-29
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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 $a=5^2-8 \cdot 5+10=-5\,$.

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    Wish the downvoter had left a comment why. – 2017-01-29