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I have been trying this question for a long time but I am not getting it. So please help me and try to make it as fast as possible

JimmyK4542
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Anjana M R
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  • Use https://en.wikipedia.org/wiki/Long_division – lab bhattacharjee Jun 21 '16 at 05:32
  • I do not think this comment is particular useful. The insight of this exercise should be that you have to look at the equation $f(i)=0$ and compare real and imaginary part to get $a$ and $b$. – MooS Jun 21 '16 at 05:42

4 Answers4

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Long division gives $$ \frac{x^4+x^3+8x^2+ax+b}{x^2+1}=x^2+x+7+\frac{(a-1)x+b-7}{x^2+1} $$ Therefore, the remainder is zero iff $(a-1)x+b-7$ is the zero polynomial, that is, iff $$ .\begin{cases} a-1=0\\ b-7=0\\ \end{cases} $$ Hence, for $a=1$ and $b=7$, we have that $x^4+x^3+8x^2+ax+b$ is divisible by $x^2+1$.

boaz
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Hint: If $x^4+x^3+8x^2+ax+b$ is divisible by $x^2+1$, then there is a polynomial $Q(x)$ such that $$x^4+x^3+8x^2+ax+b = (x^2+1)Q(x)$$ for all complex numbers $x$.

Now plug in $x = i$ and $x = -i$ and see what you get.

JimmyK4542
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    Why $i$ and $-i$? The resulting equations will be equivalent. – MooS Jun 21 '16 at 05:41
  • I think you have not understood my comment. I said that it is sufficient to plug in $i$ or $-i$, since the resulting equations will be equivalent... – MooS Jun 21 '16 at 06:20
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    @MooS, plugging in both relieves you of the obligation to prove that $a$ and $b$ can only be real. – Barry Cipra Jun 21 '16 at 13:37
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You simply write out a skeleton $(x^4+x^3+8x^2+ax+b)=(x^2+1)(cx^2+dx+e)$.

(1) comparing the $x^4$ term we must have $c=1$.

(2) comparing the $x^3$ term we have $d=1$.

(3) comparing the constant term $e=b$.

So we now have $(x^4+x^3+8x^2+ax+b)=(x^2+1)(x^2+x+b)$.

(4) comparing the $x^2$ term $8=b+1$, so $b=7$.

(5) comparing the $x$ term $a=1$.

Check: $x^4+x^3+8x^2+x+7=(x^2+1)(x^2+x+7)$.

With practice a lot of this is done in your head and it is all quite fast.

almagest
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Method 1.

Subtracting $x^2(x^2+1)$ from $x^4+x^3+8x^2+a x+b$ gives $x^3+7 x^2 +a x+b.$

Subtracting $x(x^2+1)$ from $x^3+7 x^2+a x+b$ gives $7 x^2+(a-1)x+b.$

Subtracting $7(x^2+1)$ from $7 x^2+(a-1)x+b$ gives $(a-1)x+(b-7).$

Observe that $(a-1)x+(b-7)=0$ for every $x$ if and only if $a=1$ and $b=7.$

Method 2.

$f(x)=x^4+x^3+8 x^2+a x+b$ is divisible by $x^2+1$ if $f(i)=f(-i)=0,$ where $i^2+1=0.$

We have $i^3=i(i^2)=i(-1)=-i$ , and $i^4=(i^2)^2=(-1)^2=1.$

So $f(i)=(b-7)+i(a-1)$ and $f(-i)=(b-7)-i(a-1).$

These are both $0$ if and only if $b-7=a-1=0.$

DanielWainfleet
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