List all possible (or potential) rational zeros for the polynomial below. Find all real zeros of the polynomial below and factor completely over the real numbers. Please show all of your work.
f(x) = x^4 - 7x^3 -3x^2 + 19x + 14
Help
trying a few factors of 14, I found
f(-1) = 0 and f(2) = 0
so after 2 consecutive synthetic divisions, that left me with
(x+1)(x-2)((x^2 - 6x - 7)
= (x+1)(x-2)(x+1)(x-7)
so we have a double zero at x=-1
and two more zeros at 2 and 7
I just wasted my valuable time doing the above question before noticing that your same question had already been answered.
Do not repost the same question without checking if it has been answered.
To find the potential rational zeros of the polynomial f(x) = x^4 - 7x^3 - 3x^2 + 19x + 14, we can utilize the Rational Root Theorem. According to this theorem, the possible rational zeros (or potential rational zeros) of a polynomial are all the numbers that can be obtained by dividing a factor of the constant term by a factor of the leading coefficient.
In this case, the leading coefficient is 1, and the constant term is 14. The factors of 14 are 1, 2, 7, and 14, and the factors of 1 are 1 and -1. Using these factors, you can form the potential rational zeros:
±1, ±2, ±7, ±14
Now let's find the real zeros and factor the polynomial:
To find the real zeros, we can use a combination of synthetic division, factoring, and the quadratic formula. We'll start with the potential rational zeros.
1. Try x = 1:
- Using synthetic division, we divide f(x) by (x - 1):
| 1 | -7 | -3 | 19 | 14
| 1 | -6 | -9 | 10 |
------------------------------------
1 | -6 | -9 | 10 | 24
- The remainder is 24, indicating that x = 1 is not a zero.
2. Try x = -1:
- Using synthetic division, we divide f(x) by (x + 1):
| 1 | -7 | -3 | 19 | 14
| 1 | -8 | 5 | -14 |
------------------------------------
1 | -8 | -3 | 24 | 0
- The remainder is 0, indicating that x = -1 is a zero.
3. Now that we have found one zero, which is x = -1, we can factor out (x + 1):
- The polynomial can be rewritten as f(x) = (x + 1)(x^3 - 8x^2 - 3x + 14)
4. Next, let's find the zeros of the cubic polynomial x^3 - 8x^2 - 3x + 14:
- We can use synthetic division again and test the potential rational zeros.
5. Try x = 2:
- Using synthetic division, we divide the cubic polynomial by (x - 2):
| 2 | -8 | -3 | 14
| 2 | -12 | 18 | 14 |
------------------------------------
2 | -6 | 15 | 32
- The remainder is 32, indicating that x = 2 is not a zero.
6. Try x = -2:
- Using synthetic division, we divide the cubic polynomial by (x + 2):
| -2 | -8 | -3 | 14
| -2 | 10 | -14 | 0 |
------------------------------------
-2 | 2 | 7 | 0
- The remainder is 0, indicating that x = -2 is a zero.
7. Now we can factor out (x + 2) from the cubic polynomial:
- The polynomial can be rewritten as f(x) = (x + 1)(x + 2)(x^2 - 6x + 7)
8. Factor the quadratic x^2 - 6x + 7:
- This quadratic cannot be easily factored. We'll use the quadratic formula to find the remaining zeros.
- Using the quadratic formula (x = [-b ± √(b^2 - 4ac)] / 2a), we have:
x = [6 ± √((-6)^2 - 4(1)(7))] / (2)(1)
x = [6 ± √(36 - 28)] / 2
x = [6 ± √8] / 2
x = [6 ± 2√2] / 2
x = 3 ± √2
- The remaining zeros are x = 3 + √2 and x = 3 - √2.
Therefore, the real zeros of the polynomial f(x) = x^4 - 7x^3 - 3x^2 + 19x + 14 are:
x = -1, -2, 3 + √2, and 3 - √2.
The polynomial is completely factored over the real numbers as:
f(x) = (x + 1)(x + 2)(x - (3 + √2))(x - (3 - √2))
Please note that the steps provided above offer a clear method for finding the rational zeros, real zeros, and factoring the polynomial.