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The same question was posed here Find the value of a+b+c but I cannot make sense of the answer.

My long division must be incorrect as I ended up with $a+b+c$ which is definitely wrong. I added $0$ to represent $x^3$ and $x^1$ terms.

I know the remainder should be zero if $x+1$ is a factor.

I am not putting two and two together.

saulspatz
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user554754
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    If $x+1$ is a factor then $x=-1$ is a root. @user554754 can you finish? – Minz Apr 26 '18 at 03:30
  • You have to put $ signs around your MathJax for it to be formatted. – saulspatz Apr 26 '18 at 03:31
  • In fact, if a definite answer exists, only one answer i spossible – Hagen von Eitzen Apr 26 '18 at 03:37
  • Your result is $ax^4+bx^2+c=(\cdots)\cdot(x+1)+\color{red}{a+b+c}$ and you know that it should be $ax^4+bx^2+c=(\cdots)\cdot(x+1)+\color{red}{0}$ and the question asks for the value of $a+b+c$. The rest is really like puting two and two together – Hagen von Eitzen Apr 26 '18 at 03:40
  • I added a few details to the linked question. – Ross Millikan Apr 26 '18 at 03:43
  • My long division must be incorrect as I ended up with $a+b+c\;$ **which is definitely wrong** Why wrong? That is in fact the correct remainder:

    $$ \begin{align} ax^4 + bx^2 + c &= ax^3(x \color{red}{+1}) \color{red}{- ax^3} + bx^2 + c \ &= ax^3(x+1) -ax^2(x\color{red}{+1}) \color{red}{+ax^2} + bx^2 + c \ &= (ax^3-ax^2)(x+1)+(a+b)x(x\color{red}{+1}) \color{red}{-(a+b)x}+c \ &= \big(ax^3-ax^2+(a+b)x\big)(x+1) -(a+b)(x\color{red}{+1})\color{red}{+a+b}+c \ &= \big(ax^3-ax^2+(a+b)x-(a+b)\big)(x+1) + \boxed{a+b+c} \end{align} $$

     – dxiv Apr 26 '18 at 03:52
  • Oh wow I misread the question. So its asking for the overall value of a+b+c which is equal to zero? I thought I had to find the value of a, b, and c which seems impossible with no coefficient. – user554754 Apr 26 '18 at 13:55

1 Answers1

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$x = -1$ is a root of polynomial
put $x = -1$ in $a x^4 + b x^2 + c$
$a + b + c = 0$

kayush
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