Polynomials, what are the values of "m" and "n"?

Question

The equation $x³+mx²+2x+n=0$,where $m$ and $n$ are real numbers, admits $1 + i$ as root. What are the values of $m$ and $n$?

NOTE: I developed from Viète's formulas, what are the values of $m$ and $n$?

Answer 1

If $k$ is a root then $P(k) =0$ so first thing to do is figure

$(1+i)^3 + m(1+i)^2 + 2(1+i)+n= 0$

$(1+3i + 3i^2 +i^3) + m(1+2i+ i^2) + 2(1+i) + n = 0$

$(-2 + 2i) +m(2i) + (2+2i) + n = 0$

$n + (2m+4)i = 0$

.... well, I wasn't expecting it to be so easy.

As $n, (2m+4)$ are real we must have $n=0$ and $2m+4 = 0$ so $m=-2$.

I was expecting us to get some unresolvable expression of $m$ and $n$.

Then we'd need to know that if $a+bi$ is a root of a real polynomial that $(a-bi)$ is a root as well and calculate

$(1-i)^3 + m(1-i)^2 + 2(1-i) + n=0$

And get a second expression of $m$ and $n$ which as we have two linear equations would be solvable together.

Indeed

$(1-i)^3 + m(1+i)^2 + 2(1+i) + n = 0$

$(1 - 3i +3i^2 - i^3) + m(1 - 2i + i^2) + 2(1-i) + n =0$

$(-2 -2i) + m(-2i) + (2-2i) + n=0$

$n- (2m+4)i = 0$

So, again, $n = 0$ and $m=-2$.

Answer 2

Since $1+i$ is a root and since the coefficients are real, $1-i\left(=\overline{1+i}\right)$ is another root. Let $r$ be the third root. Then the sum of the roots is $2+r=-m$. And the product of the roots is $2r=-n$. Finally,$$(1+i)(1-i)+r(1+i)+r(1-i)=2+2r=2.$$So, $r=0$ and therefore $m=-2$ and $n=0$.

Answer 3

Here are some ideas which simplify the arithmetic somewhat for the method of direct substitution.

Note that $(1+i)^2=2i$ so we can simplify things a little when we substitute $x=1+i$. Also in equating real and imaginary parts we expect two equations in two unknowns. Note, if you know the rules for the argument of a product you will know that this will come out pure imaginary before you do the calculation.

So we get$$2i(1+i)+2im+2+2i+n=0$$

and we notice that the coefficient of $m$ is pure imaginary (which was predictable from the observations above) and the coefficient of $n$ is real.

Equating real parts we get $n=0$ and imaginary parts give $4i+2im=0$ whence $m=-2$

Answer 4

By here $\,w= 1\!+\!i\,$ a root of $\,f(x)\in\Bbb R[x]\,\Rightarrow$ so too is its conjugate $\,\bar w = 1\!-\!i\,$ thus

$\begin{align} f(x) = (x\!-\!w)(x\!-\!\bar w)(x\!-\!r) &= (x^2\!\color{#90f}{-\!2}x\!+\!\color{#0a0}2)(x\!+\!\color{#90f}{n/2})\\[.2em] &= x^3\!+\!mx^2\!+\!\color{#c00}2x\!+\!n\end{align}$

Comparing coefs of $\,x^1\!:\ \color{#c00}2 = \color{#0a0}2\color{#90f}{-n}\, $ so $\,n=0$
Comparing coefs of $\,x^2 = -{\rm root\ sum}\!:\ {-}m = w\!+\!\bar w\!+\!0 = 2\ $ so $\ m=-2$

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Alternatively $\,\ \color{#c00}{w^2 = 2w\!-\!2}\,\Rightarrow\, \color{#0a0}{w^3} = w(\color{#c00}{w^2}) =\color{#0a0}{2w\!-\!4}\ $ hence

$ 0\! =\! f(w) = \underbrace{\color{#0a0}{2w\!-\!4}}_{\Large\color{#0a0}{w^3}}\! +\! m(\underbrace{\color{#c00}{2w\!-\!2}}_{\Large\color{#c00}{w^2}})\! +\! 2w\!+\!n = (2m\!+\!4)w + n\!-\!(2m\!+\!4)\,\Rightarrow\,m=-2,\, n= 0$