Find all complex and real roots of higher degree polynomials, given one root

Question

$2+3i$ is a zero of $f(x)=x^4-4x^3+17x^2-16x+52$, find all of the zeros of $f(x)$

thanks!

Answer 1

Outline: Since the coefficients are real, the complex conjugate of $2+3i$ is also a zero of $f(x).

It follows that $(x-(2+3i))(x-(2-3i))$ divides $f(x)$.

Divide $f(x)$ by $x^2-4x+13$. You will get a quadratic, and you know how to find roots of quadratics.

Remark: Actually, we don't need to do all this. That is because sadly, by inspection, $2$ is a root. The sum of the roots is $4$, so the other real root is $-2$. Thus the roots are $2+3i$. $2-3i$, $2$, and $-2$.

Answer 2

Hints.

1. As $f$ has real coefficients and you are given one root, you can immediately write down another root.
2. These two roots give a real quadratic factor of $f$.
3. Division gives another quadratic factor.
4. I'm sure you can solve a quadratic.

Answer 3

Since the coefficients are all real, $(2-3i)$ will also be a root. You can then factor out $(x-2-3i)(x-2+3i)=x^2-4x+13$ from the original polynomial. This will leave another quadratic factor, for which you can find roots by the quadratic formula.

Answer 4

For a polynomial $ f(x) $ with real coefficients, if a complex number $ z $ is a root of $ f(x) = 0 $, then the conjugate $ \bar{z} $ is also a root of the equation. This is so that the properties of sum of roots and product of roots are satisfied.

Hence, you have the second root as $ 2-3i $. Assume the remaining roots to be $a, b$. You have $$ (x-a)(x-b)(x-2-3i)(x-2+3i) = x^4 - 4x^3 + 17x^2 - 16x + 52 = 0 $$

Answer 5

Let
$\quad \alpha_1 = 2+3i$
$\quad \alpha_2 = 2-3i$
$\quad \beta \;\,= a + bi$
$\quad \gamma \;\,= c +di$

be the four complex roots to the real polynomial $f(x) = x^4-4x^3+17x^2-16x+52$. We could continue by dividing polynomials, but we can go at it another way.

Simple algebra (expanding the product of linear factors) tells us that the coefficient of $x^3$ for $f(x)$ is equal to $-(Re(\alpha_1) + Re(\alpha_2)+ Re(\beta)+ Re(\gamma))= -4-a-c$, so

$\tag 1 a+c = 0$

Again, looking at the constant coefficient, we get that

$\tag 2 \alpha_1 \alpha_2 \beta \gamma = 13 \beta \gamma = 52$

Now if $\beta$ and $\gamma$ are real numbers, by (2) they are both positive or both negative. But (1) shows that is not possible. So, they must be be complex conjugates, and (1) means the real part must be zero. So taking absolute values, $13 \beta \gamma = 13 {|\beta|}^2 = 52$, so the other two roots are $2i$ and $-2i$.