how do you factor X^3+X^2-9X-9
pls help. I don't know what to do
looks like grouping might be the simplest way ...
X^3+X^2-9X-9
= x^2(x+1) - 9(x+1)
= (x+1)(x^2-9) and now I see a difference of squares
= (x+1)(x+1)(x-1)
or
= (x-1)(x+1)^2
or, noting the "9" :-)
(x-1)(x-3)(x+3)
To factor the given expression, we can use a method called synthetic division. Here's how you can do it:
Step 1: Find a possible rational root.
The Rational Root Theorem states that any rational root (if it exists) of a polynomial with integer coefficients will have the form p/q, where p is a factor of the constant term (-9 in this case) and q is a factor of the leading coefficient (1 in this case).
To find the possible rational roots, we need to find all the factors of -9 and 1: ±1, ±3, ±9.
Step 2: Test the possible root using synthetic division.
We will test the possible roots one by one by using synthetic division. Start with one of the possible roots and divide the polynomial expression by (x - root). Repeat this process for all possible roots until you find one that gives a remainder of zero.
Let's start with the possible root x = 1.
1 | 1 1 -9 -9
| 1 2 -7
-----------------
| 1 2 -7 -16
Since the remainder is -16, x = 1 is not a root.
Let's try another possible root: x = -1.
-1 | 1 1 -9 -9
| -1 0 9
-----------------
| 1 0 -9 0
-1 is a root that gives a remainder of 0. This means (x + 1) is a factor of the given polynomial.
Step 3: Divide the polynomial expression by the polynomial factor (x + 1).
By dividing the polynomial by (x + 1), we can write the original polynomial as:
(x + 1)(x^2 - 9)
Step 4: Factor the remaining quadratic expression (x^2 - 9).
The quadratic expression (x^2 - 9) is a difference of squares, so we can further factor it:
(x + 1)(x - 3)(x + 3)
Therefore, the factored form of the given polynomial expression x^3 + x^2 - 9x - 9 is (x + 1)(x - 3)(x + 3).