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Given that $a$ and $b$ are real constants, prove that $ai$ is a root of the equation


Find the other roots of the equation in terms of $a$ and $b$.

I realise the conjugate complex pair of $ai$ is another root, i.e. $-ai$ but how do I prove that $ai$ is a root in the first place and also find the third root?

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up vote 3 down vote accepted

You prove $ai$ is a root by evaluating $p(x) = x^3 - bx^2 + a^2x - a^2b$ at $x =ai$: $p(ai) = -a^3i + a^2b + a^3i - a^2 b = 0$. Thus, since we have $a, b \in \Bbb R$, $-ai$ is a root as well. Thus $(x + ai)(x - ai) = (x^2 + a^2) \vert p(x)$. Now you can use synthetic division, or make a good guess by inspection: the fact that the term of degree $0$ is $-a^2b$ provides a strong hint, as do the facts that the quotient polynomial is monic of degree $1$. How about $x - b$? Voila, we find $p(x) = (x^2 + a^2)(x - b)$! The roots are thus $\pm ai, b$.

Hope this helps. Cheers.

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One way of proving that it's a root is substituting $ai$ in for $x$ on the left hand side of the equation and seeing whether or not you come up with the right hand side.
After you see that it is satisfactory, you can divide the left hand side polynomial by: $$ (x-ai)(x+ai) = x^2 + a^2 $$ to come up with the third root.

$$ \frac{x^3-bx^2+a^2x-a^2b}{x^2+a^2} = x-b $$ Thus $x=b$ is the third root, because we have: $$ x^3-bx^2+a^2x-a^2b = (x-ai)(x+ai)(x-b) $$

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The most straightforward way to show that $ai$ is a root of the polynomial $$f(x):=x^3-bx^2+a^2x-a^2b$$ is the substitute this value in for $x$ and find that $$f(ai)=0.$$

Once you know you have a root, the complex conjugate is also a root, which is $-ai$ in this case (as you point out); see Complex conjugate root theorem.

The Fundamental Theorem of Algebra implies there will be three (not necessarily distinct) roots. So $f(x)$ can be factored $$f(x)=(x-ai)(x+ai)(x-k)$$ where $k$ is the remaining root. If we expand out the above formula we obtain $$f(x)=x \times \text{something}-a^2k.$$ Equating the coefficients gives the remaining root.

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If $\pm ai$ are roots., then $(x-ai)(x+ai)=x^2+a^2$ is a factor of your polynomial. Polynomial long division will show this (there'll be no remainder). Also, this will find the third root.

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Substitute $x=ai$ into the expression $x^3-bx^2+a^2x-a^2b$ and show that this equates to zero.

As you correctly observed, if $ai$ is a root then so is its conjugate $-ai$. Therefore two factors of this expression must be $(x-ai)$ and $(x+ai)$. Therefore $(x-ai)(x+ai)$ must also be a factor. This gives $(x^2+a^2)$ as a factor. Use this fact to help you factorise the expression and find the remaining third factor.

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There were couple answers about checking that $ai$ is a root.

The third root can be easily found by the formulas which relate the roots with the coefficients.

The sum of the roots is $-\frac{-b}{1}=b$ or the product of the roots is $-\frac{-a^2b}{1}=a^2b$.

In general, if $x_1,..,x_n$ are all the roots of $a_nx^n+....+a_1x+a_0$ we have

$$x_1+..+x_n=-\frac{a_{n-1}}{a_n}$$ $$x_1\cdot...\cdot x_n = (-1)^n\frac{a_{0}}{a_n}$$

P.S. One of those two relations is sufficient if you know all but one root. If you are missing two roots, combining those two relations yield a quadratic equation whose roots are the two missing roots.

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