Use the rationals theorem to find all the zeros of the polynomial function. Use the zeros to factor f over the real numbers.
f(x)=x^3-2x^2-13x-10
x=
f(1) = 1-2-13-10 ≠ 0
f(-1) = -1 - 2 + 13 -10 = 0 , so x+1 is a factor
by synthetic division
x^3 - 2x^2 - 13x - 10 = (x+1)(x^2 - 3x - 10)
= (x+1)(x-5)(x+2) = 0 for roots
x = -1 or x = -2 or x = 5
To find the zeros of the polynomial function f(x) = x^3 - 2x^2 - 13x - 10 using the Rational Root Theorem, we need to look for possible rational roots of the form p/q, where p is a factor of the constant term (-10) and q is a factor of the leading coefficient (1).
The factors of -10 are ±1, ±2, ±5, ±10, and the factors of 1 are ±1. Therefore, the possible rational roots are:
±1, ±2, ±5, ±10
To test each of these potential roots, we can use synthetic division or long division. I will show how to test the potential root x = 1 using synthetic division:
Step 1: Write down the coefficients of the polynomial function in descending order:
1| 1 - 2 - 13 - 10
Step 2: Bring down the leading coefficient:
1| 1 - 2 - 13 - 10
1
Step 3: Multiply the potential root (1) by the leading coefficient (1), and write the result below the next coefficient:
1| 1 - 2 - 13 - 10
1
-------
1
Step 4: Add the numbers in the second row:
1| 1 - 2 - 13 - 10
1
------
1 - 1
Step 5: Multiply the potential root (1) by the previous result (-1), and write the result below the next coefficient:
1| 1 - 2 - 13 - 10
1 - 1
------
1 - 1
Step 6: Repeat the previous steps for each subsequent coefficient:
1| 1 - 2 - 13 - 10
1 - 1 12
------
1 - 1 -1
The result of the synthetic division is a remainder of -1, which means that x = 1 is not a zero of the polynomial function f(x).
Repeat the process for each of the remaining potential rational roots: x = -1, x = 2, x = -2, x = 5, x = -5, x = 10, and x = -10.
Upon testing all the potential roots, we find that the zeros of the function are x = -2, x = -1, and x = 5.
To factor f(x) over the real numbers, we can express the polynomial as a product of its linear factors using the zeros we found:
f(x) = (x + 2)(x + 1)(x - 5)
Therefore, the factored form of f(x) over the real numbers is (x + 2)(x + 1)(x - 5).