List all possible rational zeros of the polynomial and then show that the equation has no rational root.
x^5 − 8x^3 + 6x^2 + x − 2 = 0
the only possible rational roots are ±1,±2
so, try them.
To find the possible rational zeros of the polynomial, we can use the Rational Root Theorem.
According to the theorem, the possible rational zeros are all the factors of the constant term (in this case, -2) divided by all the factors of the leading coefficient (in this case, 1).
The factors of -2 are ±1 and ±2, and the factors of 1 are ±1.
So, the possible rational zeros are ±1, ±2.
To show that the equation has no rational root, we can use the Rational Root Theorem again. We can substitute each of the possible rational zeros into the equation and check if any of them make the equation equal to zero.
Substituting x = 1 into the equation:
(1)^5 - 8(1)^3 + 6(1)^2 + (1) - 2 = 1 - 8 + 6 + 1 - 2 = -2
Substituting x = -1 into the equation:
(-1)^5 - 8(-1)^3 + 6(-1)^2 + (-1) - 2 = -1 + 8 + 6 - 1 - 2 = 10
Substituting x = 2 into the equation:
(2)^5 - 8(2)^3 + 6(2)^2 + (2) - 2 = 32 - 64 + 24 + 2 - 2 = -8
Substituting x = -2 into the equation:
(-2)^5 - 8(-2)^3 + 6(-2)^2 + (-2) - 2 = -32 + 64 + 24 - 2 - 2 = 52
From the calculations, we can see that none of the possible rational zeros (-1, 1, -2, 2) satisfy the equation. Therefore, the equation has no rational root.
To find the possible rational zeros of the polynomial, we can use the Rational Root Theorem. According to the Rational Root Theorem, the possible rational zeros of a polynomial can be found by taking the factors of the constant term (in this case, -2) and dividing them by the factors of the leading coefficient (in this case, 1).
The factors of -2 are ±1 and ±2, and the factors of 1 are ±1. Therefore, the possible rational zeros are:
±1/1, ±2/1, ±1/2, and ±2/2.
Simplifying, we get: ±1, ±2, ±1/2, and ±1.
Now, to show that the equation has no rational roots, we can use synthetic division to test each of the possible rational zeros. If none of the rational zeros result in a remainder of zero, then we can conclude that the equation has no rational roots.
Using synthetic division, we can divide the polynomial by each of the possible rational zeros:
For x = 1: (Performing synthetic division)
1 │ 1 -8 6 1 -2
└─────
1 -7 -1 -6 -8
The remainder is -8, which is not equal to zero.
For x = -1:
-1 │ 1 -8 6 1 -2
└─────
1 -9 15 -16 18
The remainder is 18, which is not equal to zero.
We can continue this process for each of the remaining possible rational zeros.
For x = 2:
2 │ 1 -8 6 1 -2
└─────
1 -6 6 13 24
The remainder is 24, which is not equal to zero.
For x = -2:
-2 │ 1 -8 6 1 -2
└─────
1 -10 26 -51 100
The remainder is 100, which is not equal to zero.
For x = 1/2:
1/2 │ 1 -8 6 1 -2
└─────
1 -7.5 7.75 4.875 3.125
The remainder is approximately 3.125, which is not equal to zero.
For x = -1/2:
-1/2 │ 1 -8 6 1 -2
└──────
1 -8 5 -3 1
The remainder is 1, which is not equal to zero.
Finally, for x = -1:
-1 │ 1 -8 6 1 -2
└─────
1 -9 15 -16 18
The remainder is 18, which is not equal to zero.
Since none of the possible rational zeros result in a remainder of zero, we can conclude that the equation x^5 − 8x^3 + 6x^2 + x − 2 = 0 has no rational roots.