Factorizing x^3-3x^2-9x-5
Step 1: Identify the Rational Roots
To factorize the given polynomial, we need to find its rational roots using the Rational Root Theorem. The theorem states that if a polynomial has a rational root, then it must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
In this case, the constant term is -5, and the leading coefficient is 1. Therefore, the possible rational roots are:
Step 2: Test the Rational Roots
We can use synthetic division to test each of the possible rational roots. If a root is indeed a root of the polynomial, then the remainder will be zero.
1 | 1 -3 -9 -5
| 1 -2 -11
|___________
1 -2 -11 -16
The remainder is not zero, so x = 1 is not a root of the polynomial.
-1 | 1 -3 -9 -5
| -1 4 5
|_________
1 -4 -5 0
The remainder is zero, so x = -1 is a root of the polynomial.
Step 3: Factorize the Polynomial
Now that we know that x = -1 is a root of the polynomial, we can use synthetic division to factorize the polynomial.
-1 | 1 -3 -9 -5
| -1 4 5
|_________
1 -4 -5 0
Using synthetic division, we get:
x^3-3x^2-9x-5 = (x + 1)(x^2 - 4x - 5)
Now we need to factorize the quadratic expression x^2 - 4x - 5. We can use the quadratic formula to find its roots:
x = (-(-4) ± sqrt((-4)^2 - 4(1)(-5))) / (2(1))
x = (4 ± sqrt(36)) / 2
x = 2 ± 3
Therefore, the roots of the quadratic expression are x = 5 and x = -1. We can use these roots to factorize the quadratic expression:
x^2 - 4x - 5 = (x - 5)(x + 1)
Step 4: Final Answer
Substituting the factors of the polynomial, we get:
x^3-3x^2-9x-5 = (