How do you find a polynomial function f with real coefficients of the indicated degree that possesses the given zeros: degree 2; 4+3i?

Jul 2, 2017

$f \left(x\right) = {x}^{2} - 8 x + 25$

Explanation:

We know that the function is a degree-2 polynomial, so the function has two solutions, one of which is given as $4 + 3 i$.

Know that if one imaginary species is a solution to a polynomial equation, its conjugate is also a solution.

Therefore, the other solution to this equation is the conjugate of $4 + 3 i$, which is color(blue)(4-3i.

Now that we know the solutions to the polynomial equation, let's derive a function, by setting them equal to zero like so:

$\left(x - \left(4 + 3 i\right)\right) \left(x - \left(4 - 3 i\right)\right) = 0$

which is the same as

$\left(x - 4 - 3 i\right) \left(x - 4 + 3 i\right) = 0$

Factoring out the polynomial gives us

${x}^{2} - 4 x + 3 x i - 4 x + 16 - 12 i - 3 x i + 12 i - 9 {i}^{2} = 0$

Combining like terms..

${x}^{2} - 8 x + 16 - 9 {i}^{2}$

${i}^{2} = - 1$:

${x}^{2} - 8 x + 25$

The equation of the function with zeros $4 \pm 3 i$ is thus

color(blue)(f(x) = x^2 - 8x + 25

Jul 2, 2017

${x}^{2} - 8 x + 25 = 0$

Explanation:

Complex roots of real valued polynomials always appear in conjugate pairs, so if one root of $f \left(x\right) = 0$ is:

$\alpha = 4 + 3 i$

Then another root is:

$\beta = 4 - 3 i$

Se we are looking for a polynomial of degree $2$, ie a quadratic with roots $\alpha$ and $\beta$.

So then:

Sum $= \alpha + \beta$
$\text{ } = \left(4 + 3 i\right) + \left(4 - 3 i\right)$
$\text{ } = 8$

Product $= \alpha \beta$
$\text{ } = \left(4 + 3 i\right) \left(4 - 3 i\right)$
$\text{ } = 16 - 9 {i}^{2}$
$\text{ } = 25$

And so the required equation is:

${x}^{2} - \left(\text{sum of roots")x + ("product of roots}\right) = 0$
$\therefore {x}^{2} - 8 x + 25 = 0$