Skip to main content

Section 7.9 Real Polynomials

If we are given a real polynomial, by which we mean a polynomial \(p(z)=a_{0}+\cdots + a_{n}z^{n}\) such that each \(a_i \in \mathbb{R}\text{,}\) then we have a bit more information about the roots:

If \(p(x)=a_0+a_{1}x+\cdots +a_{n}x^n\) with \(a_{i}\) real, and \(p(r)=0\text{,}\) then

\begin{align*} 0 =\overline{p(r)} \amp =\overline{a_0+a_{1}r+\cdots +a_{n}r^n}\\ {\color{magenta} (\overline{z+w}=\bar{z}+\bar{w})} \amp = \overline{a_0}+\overline{a_1 r}+\cdots + \overline{ a_n r^n}\\ {\color{magenta} (\overline{zw}=\bar{z}\bar{w})} \amp =\overline{a_0}+\bar{a_1} \bar{ r}+\cdots + \bar{ a_n}\bar{ r^n}\\ {\color{magenta} (\overline{a_{i}}=a_{i}\;\; \mbox{ since } \;\;a_{i} \;\; \mbox{are real} )} \amp ={a_0}+{a_1} \bar{ r}+\cdots + { a_n}\bar{ r}^n = p(\bar{r})\text{.} \end{align*}

Note: If any of the coeffients is not a real number then all bets are off! That is, we won't be able to factor into conclude that conjugates of roots are also roots.

Find the roots of \(x^4+2x^3-7x^2+2x-8\) given that one of them is \(i\text{.}\)

Recall that the roots come in conjugate pairs, and so \(-i\) is also a root, hence \((x-i)(x+i)=x^2+1\) is a factor in the above polynomial, so we can do polynomial long division to see how it factors: first, since the leading term of the above polynomial is \(x^4\text{,}\) we subtract a multiple of \(x^2+1\) that will eliminate the \(x^4\text{,}\) so we subtract \(x^2(x^2+1)\) from the polynomial to get

\begin{equation*} 2x^3-8x^2+2x-8 \end{equation*}

Now the leading term is \(2x^3\text{,}\) so we remove \(2x(x^2+1)\) from this to get

\begin{equation*} -8x^2-8=-8(x^2+1)\text{.} \end{equation*}

Finally, we can eliminate this term by subtracting \(-8(x^2+1)\text{.}\) Thus, adding together all the multiples of \((x^2+1)\) we subtracted, we get that Thus, we see that

\begin{equation*} x^4+2x^3-7x^2+2x-8=(x^2+1)(x^2+2x-8) \end{equation*}

Now we just need to solve \(x^2+2x-8=0\text{.}\) Using the quadratic formula, we find that the other two roots are \(2\) and \(-4\text{.}\) Thus, all the roots are \(\pm i, 2\text{,}\) and \(-4\text{.}\)

Notice how in that example, we started off just knowing one root and from that the polynomial collapsed and we could find the other 3, thus, even with partial information about the roots of a polynomial, we can use reasoning like this to solve for them all.