Answer :
To find the real zeros of the polynomial function [tex]\( f(x) = 5x^4 + 9x^3 - 19x^2 - 3x \)[/tex], we can follow these steps:
1. Factor out the common term:
The polynomial has a common factor of [tex]\( x \)[/tex] in each term, so we can factor it out:
[tex]\[
f(x) = x(5x^3 + 9x^2 - 19x - 3)
\][/tex]
This immediately gives us one of the zeros: [tex]\( x = 0 \)[/tex].
2. Solve the cubic polynomial:
We now need to solve the cubic equation [tex]\( 5x^3 + 9x^2 - 19x - 3 = 0 \)[/tex].
3. Check for rational roots:
We can use the Rational Root Theorem to test for possible rational roots, which would be factors of the constant term (-3) divided by factors of the leading coefficient (5). The possible rational roots are [tex]\( \pm 1, \pm 3, \pm \frac{1}{5}, \pm \frac{3}{5} \)[/tex].
4. Test each potential rational root:
Substitute each potential rational root into the cubic equation until one works. Through trial, you would find that [tex]\( x = -3 \)[/tex] is a solution.
5. Factorize the cubic polynomial using the found root:
Since [tex]\( x = -3 \)[/tex] is a root, divide the cubic polynomial by [tex]\( (x + 3) \)[/tex] using synthetic division or polynomial division to find the other factor.
6. Solve the remaining quadratic equation:
After division, you obtain the quadratic [tex]\( 5x^2 - 19x + 5 \)[/tex]. Solve this quadratic equation using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\( a = 5 \)[/tex], [tex]\( b = -19 \)[/tex], and [tex]\( c = 5 \)[/tex].
7. Apply the quadratic formula:
[tex]\[
x = \frac{-(-19) \pm \sqrt{(-19)^2 - 4 \times 5 \times 5}}{2 \times 5} = \frac{19 \pm \sqrt{361 - 100}}{10} = \frac{19 \pm \sqrt{261}}{10}
\][/tex]
Simplify:
[tex]\[
\sqrt{261} = \sqrt{(3)^2 \times 29} = 3\sqrt{29}
\][/tex]
So,
[tex]\[
x = \frac{19 \pm 3\sqrt{29}}{10}
\][/tex]
These are the solutions to the quadratic equation.
Final real zeros of the polynomial [tex]\( f(x) \)[/tex] are:
[tex]\[ x = 0, x = -3, x = \frac{3}{5} - \frac{\sqrt{14}}{5}, \text{ and } x = \frac{3}{5} + \frac{\sqrt{14}}{5} \][/tex]
1. Factor out the common term:
The polynomial has a common factor of [tex]\( x \)[/tex] in each term, so we can factor it out:
[tex]\[
f(x) = x(5x^3 + 9x^2 - 19x - 3)
\][/tex]
This immediately gives us one of the zeros: [tex]\( x = 0 \)[/tex].
2. Solve the cubic polynomial:
We now need to solve the cubic equation [tex]\( 5x^3 + 9x^2 - 19x - 3 = 0 \)[/tex].
3. Check for rational roots:
We can use the Rational Root Theorem to test for possible rational roots, which would be factors of the constant term (-3) divided by factors of the leading coefficient (5). The possible rational roots are [tex]\( \pm 1, \pm 3, \pm \frac{1}{5}, \pm \frac{3}{5} \)[/tex].
4. Test each potential rational root:
Substitute each potential rational root into the cubic equation until one works. Through trial, you would find that [tex]\( x = -3 \)[/tex] is a solution.
5. Factorize the cubic polynomial using the found root:
Since [tex]\( x = -3 \)[/tex] is a root, divide the cubic polynomial by [tex]\( (x + 3) \)[/tex] using synthetic division or polynomial division to find the other factor.
6. Solve the remaining quadratic equation:
After division, you obtain the quadratic [tex]\( 5x^2 - 19x + 5 \)[/tex]. Solve this quadratic equation using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\( a = 5 \)[/tex], [tex]\( b = -19 \)[/tex], and [tex]\( c = 5 \)[/tex].
7. Apply the quadratic formula:
[tex]\[
x = \frac{-(-19) \pm \sqrt{(-19)^2 - 4 \times 5 \times 5}}{2 \times 5} = \frac{19 \pm \sqrt{361 - 100}}{10} = \frac{19 \pm \sqrt{261}}{10}
\][/tex]
Simplify:
[tex]\[
\sqrt{261} = \sqrt{(3)^2 \times 29} = 3\sqrt{29}
\][/tex]
So,
[tex]\[
x = \frac{19 \pm 3\sqrt{29}}{10}
\][/tex]
These are the solutions to the quadratic equation.
Final real zeros of the polynomial [tex]\( f(x) \)[/tex] are:
[tex]\[ x = 0, x = -3, x = \frac{3}{5} - \frac{\sqrt{14}}{5}, \text{ and } x = \frac{3}{5} + \frac{\sqrt{14}}{5} \][/tex]
