QuestionFind the complex zeros of the polynomial $f(x)=x^{3}-15 x^{2}+79 x-145$ .

Studdy Solution

STEP 1

Assumptions1. The polynomial function is $f(x)=x^{3}-15 x^{}+79 x-145$ . We are looking for complex zeros of the function, which are the values of $x$ for which $f(x)=0$

STEP 2

To find the zeros of the function, we need to solve the equation $f(x)=0$ . This gives us the following equation $x^{}-15 x^{2}+79 x-145=0$

STEP 3

This is a cubic equation, and solving it exactly can be quite complex. However, we can start by trying to find one root using the Rational Root Theorem, which states that any rational root of the polynomial, expressed in lowest terms, has a numerator that is a factor of the constant term (-145) and a denominator that is a factor of the leading coefficient (1).

STEP 4

The factors of -145 are ±1, ±, ±29, ±145. We can test these values in the equation to see if any of them are roots.

STEP 5

After testing these values, we find that $x=5$ is a root of the equation. This means that $(x-5)$ is a factor of the polynomial.

STEP 6

We can perform polynomial division or use the factor theorem to find the other factor of the polynomial. The factor theorem states that if $(x-a)$ is a factor of a polynomial $f(x)$ , then $f(x)$ can be expressed as $(x-a)g(x)$ , where $g(x)$ is another polynomial.

STEP 7

Dividing the given polynomial by $(x-5)$ , we get $f(x) = (x-5)(x^{2}-10x+29)$

STEP 8

Now we need to find the roots of the quadratic equation $x^{2}-10x+29=0$ . We can use the quadratic formula to do this $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

STEP 9

Substitute $a=$ , $b=-$ , and $c=29$ into the quadratic formula $x = \frac{-(-) \pm \sqrt{(-)^{2}-4()(29)}}{2()}$

STEP 10

implify the expression under the square root $x = \frac{10 \pm \sqrt{100-116}}{2}$

STEP 11

implify further $x = \frac{10 \pm \sqrt{-16}}{}$

STEP 12

The square root of a negative number is imaginary, so we can write $\sqrt{-16}$ as $4i$ . This gives us $x = \frac{10 \pm4i}{2}$

STEP 13

implify to get the final answer $x =5 \pm2i$ So the complex zeros of the function are $5$ , $5+2i$ , and $5-2i$ .

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