3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$
    – Frank
    Jan 29 '17 at 2:03
  • $\begingroup$ @Frank, Could You please show me a bit more? I could not get. $\endgroup$
    – pi-π
    Jan 29 '17 at 2:07
4
$\begingroup$

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 $a$.

$\endgroup$
15
  • 1
    $\begingroup$ Re formatting: not exactly my cup of tea, but there are a couple of worked out examples here. $\endgroup$
    – dxiv
    Jan 29 '17 at 2:25
  • $\begingroup$ 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 $k=5$. $\endgroup$ Jan 29 '17 at 2:26
  • 1
    $\begingroup$ @OscarLanzi That was a little mistake on my part. It should be fixed. $\endgroup$
    – Frank
    Jan 29 '17 at 2:27
  • 1
    $\begingroup$ @dxiv I think I have it formatted in $\LaTeX$. :) $\endgroup$
    – Frank
    Jan 29 '17 at 2:27
  • 1
    $\begingroup$ Looks great, +1. $\endgroup$ Jan 29 '17 at 10:49
1
$\begingroup$

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 $k^2-8k+10=a\longrightarrow a=-5$.

$\endgroup$
1
$\begingroup$

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 $a=5^2-8 \cdot 5+10=-5\,$.

$\endgroup$
1
  • $\begingroup$ Wish the downvoter had left a comment why. $\endgroup$
    – dxiv
    Jan 29 '17 at 2:45
1
$\begingroup$

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,

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

$\endgroup$
3
  • $\begingroup$ Where did $b$ come from? $\endgroup$
    – pi-π
    Jan 29 '17 at 2:18
  • $\begingroup$ $b$ and $c$ are coefficient of the quotient. $\endgroup$
    – Arnaldo
    Jan 29 '17 at 2:20
  • $\begingroup$ 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? $\endgroup$
    – Arnaldo
    Jan 29 '17 at 2:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.