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If $x^3+x^2-x+1=D(x) (x^2 + 2x+1)+2$ then find $D(x)$

My Attempt,

Comparing the given expression with division algorithm we get, Dividend $=x^3+x^2-x+1$

Divisior $=D(x)$

Quotient $=(x^2+2x+1)=(x+1)^2$

Remainder $=2$.

After this I tried to use synthetic division method but could not..

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    You are essentially done, what's not working? $$D(x)=\frac{x^3+x^2-x-1}{(x+1)^2}= \cdots$$ – 2017-01-28
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    You're given the equation $P = DQ + R$; why not just solve for $D$? – 2017-01-28
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    Yeah, you should have get $D(x)=x-1$,as pjs36 and dxiv suggested – 2017-01-28

4 Answers 4

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Instead of using polynomial division, you can also compare coefficients. Namely, note that $D(x)$ must be of degree one. So write $D(x) = ax + b$ and compare

$$ x^3 + x^2 - x + 1 = (ax + b)(x^2 + 2x + 1) + 2 = ax^3 + (2a + b)x^2 + (a + 2b)x + (2 + b).$$

Hence, $2 + b = 1$ so $b = -1$ and $a + 2b = a - 2 = -1$ so $a = 1$.

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    You can also immediately see that $a = 1$ so you can set up an equation for $b$ only. – 2017-01-28
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Synthetic division is geared for division of a polynomial by $x+c.$ But in your case, since $x^2+2x+1=(x+1)^2,$ you could use synthetic division twice to do this problem, only it takes some extra work to keep track of the quotient and remainder. "Long division" is not that long in this case and might really be easier.

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In this case you can subtract $2$ from both side and solve for $D(x)$

\begin{eqnarray} D(x)&=&\frac{x^3+x^2-x-1}{x^2+2x+1}\\ &=&\frac{x^2(x+1)-(x+1)}{(x+1)(x+1)}\\ &=&\frac{x^2-1}{x+1}\\ &=&\frac{(x+1)(x-1)}{x+1}\\ &=&x-1 \end{eqnarray}

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Let $f(x)=x^3+x^2-x+1$.

Hence, $f(-1)=2$, $f'(-1)=0$, $f''(-1)=-4$ and since $D(x)=x+a$, the rest is smooth.