Question

find all zeros of the polynomial function. use the rational zero theorem, descartess rule of signs, and possibly the graph of the polynomial function shown by a graphing utility as an aid in obtaining the first zero or the first root. f(x)=x^4 - 14x^3 + 34x^2 + 114x + 65 find the zeros of the polynomial function. (simplify your answer. type an exact answer, using radicals and i as needed. use a comma to separate answers as needed. type each answer only once.)

Explanation:

Step1: Apply Rational Zero Theorem

The Rational Zero Theorem states that if a polynomial function $f(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0$ has integer coefficients, then the possible rational zeros are of the form $\frac{p}{q}$, where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading - coefficient $a_n$. For $f(x)=x^{4}-14x^{3}+34x^{2}+114x + 65$, $a_n = 1$ and $a_0=65$. The factors of $65$ are $\pm1,\pm5,\pm13,\pm65$ and the factors of $1$ are $\pm1$. So the possible rational zeros are $\pm1,\pm5,\pm13,\pm65$.

Step2: Use Descartes's Rule of Signs

For $f(x)=x^{4}-14x^{3}+34x^{2}+114x + 65$, the number of sign - changes in $f(x)$ is $2$. So there are either $2$ or $0$ positive real zeros. For $f(-x)=(-x)^{4}-14(-x)^{3}+34(-x)^{2}+114(-x)+65=x^{4}+14x^{3}+34x^{2}-114x + 65$, the number of sign - changes is $2$. So there are either $2$ or $0$ negative real zeros.

Step3: Test possible rational zeros

Test $x=-1$: $f(-1)=(-1)^{4}-14(-1)^{3}+34(-1)^{2}+114(-1)+65=1 + 14+34-114 + 65=0$. So $x=-1$ is a zero, and $(x + 1)$ is a factor of $f(x)$.

Step4: Perform polynomial long - division

Divide $f(x)=x^{4}-14x^{3}+34x^{2}+114x + 65$ by $(x + 1)$. Using polynomial long - division or synthetic division, we get $f(x)=(x + 1)(x^{3}-15x^{2}+49x+65)$.

Step5: Test for another zero of the cubic polynomial

Test $x=-1$ again for the cubic polynomial $g(x)=x^{3}-15x^{2}+49x + 65$. $g(-1)=(-1)^{3}-15(-1)^{2}+49(-1)+65=-1-15 - 49+65=0$. So $(x + 1)$ is a factor of $x^{3}-15x^{2}+49x + 65$.

Step6: Perform polynomial long - division again

Divide $x^{3}-15x^{2}+49x + 65$ by $(x + 1)$ to get $x^{3}-15x^{2}+49x + 65=(x + 1)(x^{2}-16x+65)$.

Step7: Solve the quadratic equation

Set $x^{2}-16x + 65=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-16$, and $c = 65$. $x=\frac{16\pm\sqrt{(-16)^{2}-4\times1\times65}}{2\times1}=\frac{16\pm\sqrt{256 - 260}}{2}=\frac{16\pm\sqrt{-4}}{2}=\frac{16\pm2i}{2}=8\pm i$.

Answer:

$-1,8 + i,8 - i$