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Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$ is a root of the cubic.

My effort

Working backwards I let $P(x)$ be a polynomial with roots $a,a^2+2a-14$ and $r$. Thus, $$P(x)=(x-a)(x-r)(x-(a^2+2a-14))$$

Expanding, I get

$$P(x) =(x^2-(a+r)x+ar)(x-(a^2+2a-14)) $$ $$P(x) =x^3+x^2[-(a^2+2a-14)-(a+r)]+x[(a+r)(a^2+2a-14)+ar]-ar(a^2+2a-14)$$ Equating coefficients of $P(x)$ with the given cubic $x^3-21x+35=0$ I have the following system of equations : \begin{array} \space (a^2+2a-14)+(a+r)&=0 \\ (a+r)(a^2+2a-14)+ar&=-21 \\ -ar(a^2+2a-14)&=35 \\ \end{array}

From the first equation I have $(a^2+2a-14) =-(a+r) $ which, substituted in the other two equations ,it yields

\begin{array} \space -(a+r)^2+ar &=-21 \\ ar(a+r) &=35 \\ \end{array}

Rearranging the second equation for $ar$ I have $ar=\cfrac{35}{(a+r)}$ which I now substitute into the first eq. to get:

\begin{array} \space -(a+r)^2+\cfrac{35}{(a+r)}&=-21 \\ -(a+r)^3+35 +21(a+r) &=0 \\ \end{array}

My problem now is that the last equation looks pretty darn close to $x^3-21x+35=0$ but some signs are not in the right place,which makes me wonder if I have made some careless mistake(I have already checked but I don't see it) or if I have left some algebraic manipulations to do.

Mr. Y
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    Why don't you just substitute $a^2+2a-14$ into the polynomial? – Oliver Jones Feb 03 '16 at 09:04
  • It looks painful to do all the algebra after that,but yes ...I haven't thought of that. – Mr. Y Feb 03 '16 at 09:09
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    Prove that $(x^2 + 2x -14)^3 -21(x^2 + 2x - 14) + 35$ is divisible by $x^3 -21x + 35$. – David Feb 03 '16 at 09:10
  • Thanks for the advice,it looks a lot simpler.I am now quite obsessed about what's wrong with this other method I've shown... – Mr. Y Feb 03 '16 at 09:14
  • I've edited my answer to show where you went wrong, hope this clears things up now ;) –  Feb 03 '16 at 09:24

2 Answers2

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It's quite a lot easier to simply plug in $x=a^2+2a-14$ into $P(x)$ and see \begin{align} x^3-21x+35&=(a^2+2a-14)^3-21(a^2+2a-14)+35\\ &=(35-21a + a^3) (-69 - 9 a + 6 a^2 + a^3)\\ &=0\cdot (-69 - 9 a + 6 a^2 + a^3)\\ &=0 \end{align} Note that factorizing isn't too hard since you already "know" that one factor will be $35-21a + a^3$.

Now where did you go wrong? You had $(a+r)^3-21(a+r)-35=0$. This indeed looks very much alike $P(x)$, but not quite. You didn't make any mistake in your algebra, you assumed $a+r$ would be a root of $P$, which it is not (necessarily). But, when writing \begin{align} Q(x)&=x^3+px^2+qx+r\\ &=(x-\alpha)(x-\beta)(x-\gamma)\\ &=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma \end{align} then you'll see, when $p$, the coefficient for $x^2$, is $0$, (which it is, in your case of $P(x)$), then the roots add up to $\alpha+\beta+\gamma=-p=0$, so $a+r$ is the negative of a root! Now it becomes clear that $(a+r)^3-21(a+r)-35=0$, since we know $(-(a+r))^3-21(-(a+r))+35=0$.
  • I see that.My bad,I always seem to take the peasant's way ... – Mr. Y Feb 03 '16 at 09:13
  • I'll edit my answer to include your mistake –  Feb 03 '16 at 09:14
  • Forgive me ,but can you elaborate a bit on why the root is actually $-(a+r)$ from $a=0$ (also why can you make $a=0$? this would lead to $0=35$ from the system of equations above)? I see that the solution was already there if only I had substituted back $a+r=-(a^2+20-14)$,from this last equation I see that $a+r$ is actually negative but I can't see it from your point of view. – Mr. Y Feb 03 '16 at 10:13
  • Exuse me. I had chosen my variables so that there were two different variables called $a$. I renamed the variables, so it should be clearer now. –  Feb 03 '16 at 10:16
  • No problem.I thought I was missing some part.Thanks for your answer :). – Mr. Y Feb 03 '16 at 10:17
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Since the term in $x^2$ is missing, the sum of the three roots is zero; so $a^2+2a-14$ is a root if and only if $-a-(a^2+2a-14)=-a^2-3a+14$ is also a root.

Since $$ (x-a^2-2a+14)(x+a^2+3a-14)=x^2+ax-a^4 - 5a^3 + 22a^2 + 70a - 196 $$ and the remainder of $-t^4 - 5t^3 + 22t^2 + 70t - 196$ divided by $t^3-21t+35$ is $t^2-21$, we have $$ (x-a^2-2a+14)(x+a^2+3a-14)=x^2+ax+a^2-21 $$ so \begin{align} (x-a)(x-a^2-2a+14)(x+a^2+3a-14) &=(x-a)(x^2+ax+a^2-21)\\ &=x^3-21x-a^3+21a\\ &=x^3-21x+35 \end{align} is the required factorization.

egreg
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