use the rational zeros theorem to find all the real zeros of the polynomial function,. use the zeros to factor f over the real numbers.
f(x)=11x^3-x^2+11x-1
please show all work
Hint, your order of steps should look like this:
Synthetic division
Quotient
Quotient
Solve
All Real Zeros.
To find the rational zeros of a polynomial function, we can use the rational zeros theorem. According to the theorem, any rational zero of a polynomial function can be expressed in the form p/q, where p is a factor of the constant term (in this case -1) and q is a factor of the leading coefficient (in this case 11).
In this case, the constant term is -1 and the leading coefficient is 11. Therefore, the possible rational zeros can be calculated by finding the factors of -1 and 11.
The factors of -1 are ±1, and the factors of 11 are ±1 and ±11. Therefore, the possible rational zeros are:
±1, ±1/11
Now we can use these possible rational zeros to test and determine the actual rational zeros. We can use synthetic division to test each candidate until we find one that produces a zero remainder.
Let's test x = 1:
1 | 11 -1 11 -1
| 11 10 21
---------------------
| 11 10 21 20
Since the remainder is not zero, x = 1 is not a zero of the function.
Let's test x = -1:
-1 | 11 -1 11 -1
| -11 12 -23
---------------------
| 0 -1 -12 -24
Since the remainder is zero, x = -1 is a zero of the function.
Next, let's test x = 1/11:
1/11 | 11 -1 11 -1
| 11/11 0 11/11
----------------------
| 11 10 11/11 10/11
Since the remainder is not zero, x = 1/11 is not a zero of the function.
Now that we have found x = -1 as a zero, we can factor the polynomial. We divide the polynomial function by (x - (-1)):
(x + 1)(11x^2 - 12x - 24)
This gives us the factored form of the polynomial function:
f(x) = (x + 1)(11x^2 - 12x - 24)
Therefore, the real zeros of the polynomial function f(x) = 11x^3 - x^2 + 11x - 1 are -1, and the factored form of the polynomial over the real numbers is (x + 1)(11x^2 - 12x - 24).