Answer :
To find the zeros of the polynomial function [tex]\( f(x)=x^4-2x^3-23x^2+40x+60 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make [tex]\( f(x) = 0 \)[/tex]. Here are the steps to find these zeros:
1. Set the polynomial equal to zero:
[tex]\[
x^4 - 2x^3 - 23x^2 + 40x + 60 = 0
\][/tex]
2. Identify possible rational roots:
By the Rational Root Theorem, any rational root of the polynomial equation is a factor of the constant term (60) divided by a factor of the leading coefficient (1). Possible rational roots are:
[tex]\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60
\][/tex]
3. Test the possible rational roots:
Substitute these values into the polynomial to see which ones satisfy the equation.
After testing, it turns out that [tex]\( x = -1 \)[/tex] and [tex]\( x = 3 \)[/tex] are roots of the polynomial.
4. Factorize the polynomial using these roots:
Since [tex]\( x = -1 \)[/tex] and [tex]\( x = 3 \)[/tex] are roots, [tex]\( (x + 1) \)[/tex] and [tex]\( (x - 3) \)[/tex] are factors of the polynomial. We then divide the polynomial by [tex]\( (x + 1)(x - 3) \)[/tex].
Performing the polynomial division, we would get the quotient:
[tex]\[
x^2 - 2x - 20
\][/tex]
5. Solve the remaining quadratic equation:
Now, we factorize [tex]\( x^2 - 2x - 20 \)[/tex]:
[tex]\[
x^2 - 2x - 20 = 0
\][/tex]
The solutions to this quadratic equation are found using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{2 \pm \sqrt{4 + 80}}{2}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{84}}{2}
\][/tex]
Simplify further:
[tex]\[
x = \frac{2 \pm 2\sqrt{21}}{2}
\][/tex]
[tex]\[
x = 1 \pm \sqrt{21}
\][/tex]
6. Combine all roots:
Therefore, the zeros of the polynomial function [tex]\( f(x) \)[/tex] are:
[tex]\[
x = -1, \quad x = 3, \quad x = -2\sqrt{5}, \quad x = 2\sqrt{5}
\][/tex]
These are the solutions for [tex]\( x \)[/tex] that make the polynomial equal to zero.
1. Set the polynomial equal to zero:
[tex]\[
x^4 - 2x^3 - 23x^2 + 40x + 60 = 0
\][/tex]
2. Identify possible rational roots:
By the Rational Root Theorem, any rational root of the polynomial equation is a factor of the constant term (60) divided by a factor of the leading coefficient (1). Possible rational roots are:
[tex]\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60
\][/tex]
3. Test the possible rational roots:
Substitute these values into the polynomial to see which ones satisfy the equation.
After testing, it turns out that [tex]\( x = -1 \)[/tex] and [tex]\( x = 3 \)[/tex] are roots of the polynomial.
4. Factorize the polynomial using these roots:
Since [tex]\( x = -1 \)[/tex] and [tex]\( x = 3 \)[/tex] are roots, [tex]\( (x + 1) \)[/tex] and [tex]\( (x - 3) \)[/tex] are factors of the polynomial. We then divide the polynomial by [tex]\( (x + 1)(x - 3) \)[/tex].
Performing the polynomial division, we would get the quotient:
[tex]\[
x^2 - 2x - 20
\][/tex]
5. Solve the remaining quadratic equation:
Now, we factorize [tex]\( x^2 - 2x - 20 \)[/tex]:
[tex]\[
x^2 - 2x - 20 = 0
\][/tex]
The solutions to this quadratic equation are found using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{2 \pm \sqrt{4 + 80}}{2}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{84}}{2}
\][/tex]
Simplify further:
[tex]\[
x = \frac{2 \pm 2\sqrt{21}}{2}
\][/tex]
[tex]\[
x = 1 \pm \sqrt{21}
\][/tex]
6. Combine all roots:
Therefore, the zeros of the polynomial function [tex]\( f(x) \)[/tex] are:
[tex]\[
x = -1, \quad x = 3, \quad x = -2\sqrt{5}, \quad x = 2\sqrt{5}
\][/tex]
These are the solutions for [tex]\( x \)[/tex] that make the polynomial equal to zero.
