Factor f(x)=x^4+x^3-8x^2+6x+36 completely
If there is a rational root, it would be +or- 1,2,3,4,6,8 or 12.
x = -2 works, so x+2 is a factor.
So, after one factoring step, and a bit of polynomial long division, you have
(x+2)(x^3 -x^2 -6x +18) = 0
Next, use the rational root theorm on the cubic factor. You will find that -3 is a root, making (x+3) a factor.
So now you have
x^4+x^3-8x^2+6x+36
(x+2)(x+3)(x^2 -4x +6)
The remaining two factors are complex conjugates.
To factor the expression f(x) = x^4 + x^3 - 8x^2 + 6x + 36 completely, you can follow these steps:
Step 1: Look for any common factors
Check if there are any common factors among the coefficients of the terms. In this case, all the coefficients are different, so there are no common factors other than 1.
Step 2: Look for rational roots using the Rational Root Theorem
The Rational Root Theorem states that any rational root of a polynomial equation is of the form p/q, where p is a factor of the constant term (in this case, 36) and q is a factor of the leading coefficient (in this case, 1).
To find the factors of 36, we have: ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.
To find the factors of 1, we have: ±1.
So, the possible rational roots are: ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.
Step 3: Test the possible rational roots using synthetic division
Using synthetic division, we can test each possible root until we find one that gives a remainder of 0.
Let's start by testing x = 1:
1 | 1 1 -8 6 36
| 1 2 -6 0
|-------------
1 2 -6 0
Since the remainder is 0, we found a root at x = 1. The quotient after dividing is 1x^3 + 2x^2 - 6x + 0, or simply x^3 + 2x^2 - 6x.
Step 4: Continue factoring using the quotient from step 3
Now we have x^3 + 2x^2 - 6x left to factor. Repeat the process by looking for possible rational roots and testing each using synthetic division.
Let's test x = -1:
-1 | 1 2 -6
| -1 -1 7
|-------------
1 1 -7
No remainder of 0, so x = -1 is not a root.
Let's test x = 3:
3 | 1 2 -6
| 3 15 27
|-------------
1 5 21
No remainder of 0, so x = 3 is not a root.
Let's test x = -3:
-3 | 1 2 -6
| -3 3 9
|-------------
1 -1 3
No remainder of 0, so x = -3 is not a root.
Let's test x = 2:
2 | 1 2 -6
| 2 8 4
|-------------
1 4 -2
No remainder of 0, so x = 2 is not a root.
Let's test x = -2:
-2 | 1 2 -6
| -2 -8 28
|-------------
1 0 22
No remainder of 0, so x = -2 is not a root.
Step 5: Continue testing roots using factors of the constant term
Since we have already tested all the possible rational roots, we can try using factors of the constant term (36) as possible roots.
Let's test x = 4:
4 | 1 2 -6
| 4 24 72
|-------------
1 6 66
No remainder of 0, so x = 4 is not a root.
Let's test x = -4:
-4 | 1 2 -6
| -4 8 -2
|-------------
1 -2 -8
No remainder of 0, so x = -4 is not a root.
Step 6: Write the factored form
Since we have tested all possible rational roots and none had a remainder of 0, the polynomial cannot be factored further using rational roots. Therefore, the fully factored form of f(x) = x^4 + x^3 - 8x^2 + 6x + 36 is (x - 1)(x^3 + 2x^2 - 6x).