Answer :
To find the zeros of the function [tex]\(f(x)=\frac{6x^3-19x^2+16x-4}{x+1}\)[/tex], we need to first understand that zeros of the function are values of [tex]\(x\)[/tex] for which the function equals zero.
1. Factorization Approach:
Since the function is a rational expression, we begin by dividing the polynomial in the numerator by the polynomial [tex]\(x+1\)[/tex] in the denominator. This process is known as polynomial division.
2. Perform Polynomial Division:
- Divide the polynomial [tex]\(6x^3 - 19x^2 + 16x - 4\)[/tex] by [tex]\(x + 1\)[/tex].
- After dividing, we find that we get a quotient polynomial and a remainder.
The result of this division is:
- Quotient: [tex]\(6x^2 - 25x + 41\)[/tex]
- Remainder: [tex]\(-45\)[/tex]
Now the function can be expressed as:
[tex]\[
f(x) = (x+1) \times \left(6x^2 - 25x + 41\right) + (-45)
\][/tex]
3. Finding the Zeros:
To find the zeros of [tex]\(f(x)\)[/tex]:
- Set [tex]\(f(x) = 0\)[/tex].
- This implies:
[tex]\[
(x+1)(6x^2 - 25x + 41) - 45 = 0
\][/tex]
- For the rational function to be zero, the numerator must be zero: [tex]\((x+1)(6x^2 - 25x + 41) = 0\)[/tex].
This implies:
- Either [tex]\(x+1 = 0\)[/tex] or [tex]\(6x^2 - 25x + 41 = 0\)[/tex].
- Solving [tex]\(x+1 = 0\)[/tex] gives:
- [tex]\(x = -1\)[/tex]
- Solving the quadratic equation [tex]\(6x^2 - 25x + 41 = 0\)[/tex] using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 6\)[/tex], [tex]\(b = -25\)[/tex], and [tex]\(c = 41\)[/tex], gives:
[tex]\[
x = \frac{25 \pm \sqrt{(-25)^2 - 4 \times 6 \times 41}}{2 \times 6}
\][/tex]
[tex]\[
x = \frac{25 \pm \sqrt{625 - 984}}{12}
\][/tex]
[tex]\[
x = \frac{25 \pm \sqrt{-359}}{12}
\][/tex]
- Since the discriminant ([tex]\(-359\)[/tex]) is negative, the zeros are complex:
- [tex]\(x = \frac{25}{12} - \frac{\sqrt{359}i}{12}\)[/tex]
- [tex]\(x = \frac{25}{12} + \frac{\sqrt{359}i}{12}\)[/tex]
4. Conclusion:
The zeros of the function [tex]\(f(x)\)[/tex] are [tex]\(-1\)[/tex], [tex]\(\frac{25}{12} - \frac{\sqrt{359}i}{12}\)[/tex], and [tex]\(\frac{25}{12} + \frac{\sqrt{359}i}{12}\)[/tex]. These include one real zero, [tex]\(-1\)[/tex], and two complex zeros.
1. Factorization Approach:
Since the function is a rational expression, we begin by dividing the polynomial in the numerator by the polynomial [tex]\(x+1\)[/tex] in the denominator. This process is known as polynomial division.
2. Perform Polynomial Division:
- Divide the polynomial [tex]\(6x^3 - 19x^2 + 16x - 4\)[/tex] by [tex]\(x + 1\)[/tex].
- After dividing, we find that we get a quotient polynomial and a remainder.
The result of this division is:
- Quotient: [tex]\(6x^2 - 25x + 41\)[/tex]
- Remainder: [tex]\(-45\)[/tex]
Now the function can be expressed as:
[tex]\[
f(x) = (x+1) \times \left(6x^2 - 25x + 41\right) + (-45)
\][/tex]
3. Finding the Zeros:
To find the zeros of [tex]\(f(x)\)[/tex]:
- Set [tex]\(f(x) = 0\)[/tex].
- This implies:
[tex]\[
(x+1)(6x^2 - 25x + 41) - 45 = 0
\][/tex]
- For the rational function to be zero, the numerator must be zero: [tex]\((x+1)(6x^2 - 25x + 41) = 0\)[/tex].
This implies:
- Either [tex]\(x+1 = 0\)[/tex] or [tex]\(6x^2 - 25x + 41 = 0\)[/tex].
- Solving [tex]\(x+1 = 0\)[/tex] gives:
- [tex]\(x = -1\)[/tex]
- Solving the quadratic equation [tex]\(6x^2 - 25x + 41 = 0\)[/tex] using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 6\)[/tex], [tex]\(b = -25\)[/tex], and [tex]\(c = 41\)[/tex], gives:
[tex]\[
x = \frac{25 \pm \sqrt{(-25)^2 - 4 \times 6 \times 41}}{2 \times 6}
\][/tex]
[tex]\[
x = \frac{25 \pm \sqrt{625 - 984}}{12}
\][/tex]
[tex]\[
x = \frac{25 \pm \sqrt{-359}}{12}
\][/tex]
- Since the discriminant ([tex]\(-359\)[/tex]) is negative, the zeros are complex:
- [tex]\(x = \frac{25}{12} - \frac{\sqrt{359}i}{12}\)[/tex]
- [tex]\(x = \frac{25}{12} + \frac{\sqrt{359}i}{12}\)[/tex]
4. Conclusion:
The zeros of the function [tex]\(f(x)\)[/tex] are [tex]\(-1\)[/tex], [tex]\(\frac{25}{12} - \frac{\sqrt{359}i}{12}\)[/tex], and [tex]\(\frac{25}{12} + \frac{\sqrt{359}i}{12}\)[/tex]. These include one real zero, [tex]\(-1\)[/tex], and two complex zeros.
