factor x^4-3x^3-11x^2+3x+10 completely using the factor theorem.
using the factor theorem, I tried
x = ±1, ±2, ±5
I found x=1 and x=-1 to work as well as x=-2
so factors would be
(x+1)(x-1)(x+2)(........)
I can reason that the first term of the missing factor is x and it must end in -5
so
x^4-3x^3-11x^2+3x+10
= (x-1)(x+1)(x+2)(x-5)
or
= (x^1 - 1)(x+2)(x-5)
To factor the polynomial \(x^4-3x^3-11x^2+3x+10\) completely, we can use the Factor theorem.
1. First, we need to find the possible rational roots of the polynomial. According to the Rational Root Theorem, the possible rational roots are the factors of the constant term (10) divided by the factors of the leading coefficient (1), so our possible rational roots are: ±1, ±2, ±5, and ±10.
2. We can now use synthetic division to test the possible roots. Let's start with 1:
```
1 | 1 -3 -11 3 10
| 1 -2 -13 -10
___________________
1 -2 -13 -10 0
```
Since the remainder is zero, we conclude that \(x-1\) is a factor.
3. Using the factor \(x-1\), we can write the polynomial as:
\(x^4-3x^3-11x^2+3x+10 = (x-1)(x^3-2x^2-13x-10)\)
4. Now, we need to factor the cubic polynomial \(x^3-2x^2-13x-10\). Let's try the same process as above to find another root.
- Testing -1:
```
-1 | 1 -2 -13 -10
| -1 3 10
________________
1 -3 -10 0
```
The remainder is zero, so we conclude that \(x+1\) is also a factor.
5. Using the factor \(x+1\), we can write the cubic polynomial as:
\(x^3-2x^2-13x-10 = (x+1)(x^2-3x-10)\)
6. Now, we can factor the quadratic polynomial \(x^2-3x-10\) using the quadratic formula, completing the factorization:
\(x^4-3x^3-11x^2+3x+10 = (x-1)(x+1)(x+2)(x-5)\)
So, the completely factored form of the polynomial \(x^4-3x^3-11x^2+3x+10\) is \((x-1)(x+1)(x+2)(x-5)\).
To factor the polynomial completely using the factor theorem, we need to find its roots, or values of x that would make the polynomial equal to zero. The factor theorem states that if a polynomial P(x) has a root r, then (x - r) is a factor of P(x).
Let's start by finding the possible rational roots of the given polynomial, which are the ratios of the factors of the constant term (10) to the factors of the leading coefficient (1).
Factors of 10: ±1, ±2, ±5, ±10
Factors of 1: ±1
Possible rational roots: ±{1/1, 2/1, 5/1, 10/1} = ±{1, 2, 5, 10}
Now, we can check each of these possible roots using synthetic division or substitution to see if any of them are roots of the polynomial.
Let's start with x = 1:
Substituting x = 1 into the polynomial, we get:
P(1) = (1)^4 - 3(1)^3 - 11(1)^2 + 3(1) + 10 = 1 - 3 - 11 + 3 + 10 = 0
Since P(1) = 0, we have found a root. Therefore, (x - 1) is a factor of the polynomial.
Using synthetic division, we can divide the polynomial by (x - 1) to find the quotient polynomial and the remaining factor:
1 | 1 -3 -11 3 10
| 1 -2 -13 -10
----------------------
1 -2 -13 -10 0
The quotient polynomial is x^3 - 2x^2 - 13x - 10.
Now, let's find the possible rational roots for the quotient polynomial.
Factors of 10: ±1, ±2, ±5, ±10
Factors of 1: ±1
Possible rational roots: ±{1/1, 2/1, 5/1, 10/1} = ±{1, 2, 5, 10}
Again, we can check each possible root using synthetic division or substitution.
Trying x = 1:
P(1) = (1)^3 - 2(1)^2 - 13(1) - 10 = 1 - 2 - 13 - 10 = -24 ≠ 0
Trying x = -1:
P(-1) = (-1)^3 - 2(-1)^2 - 13(-1) - 10 = -1 - 2 + 13 - 10 = 0
P(-1) = 0, so we have found another root. Therefore, (x + 1) is a factor of the quotient polynomial.
Using synthetic division, we divide the quotient polynomial by (x + 1):
-1 | 1 -2 -13 -10
| -1 3 10
-----------------
1 1 -10
The quotient polynomial is x^2 + x - 10.
Now, let's find the possible rational roots for the quadratic polynomial.
Factors of 10: ±1, ±2, ±5, ±10
Factors of 1: ±1
Possible rational roots: ±{1/1, 2/1, 5/1, 10/1} = ±{1, 2, 5, 10}
Checking each possible root:
Trying x = 1:
P(1) = (1)^2 + (1) - 10 = 1 + 1 - 10 = -8 ≠ 0
Trying x = -1:
P(-1) = (-1)^2 + (-1) - 10 = 1 - 1 - 10 = -10 ≠ 0
Trying x = 2:
P(2) = (2)^2 + (2) - 10 = 4 + 2 - 10 = -4 ≠ 0
Trying x = -2:
P(-2) = (-2)^2 + (-2) - 10 = 4 - 2 - 10 = -8 ≠ 0
Trying x = 5:
P(5) = (5)^2 + (5) - 10 = 25 + 5 - 10 = 20 ≠ 0
Trying x = -5:
P(-5) = (-5)^2 + (-5) - 10 = 25 - 5 - 10 = 10 ≠ 0
Trying x = 10:
P(10) = (10)^2 + (10) - 10 = 100 + 10 - 10 = 100 ≠ 0
Trying x = -10:
P(-10) = (-10)^2 + (-10) - 10 = 100 - 10 - 10 = 80 ≠ 0
None of the possible rational roots yield a value of 0, which means that the remaining factor is irreducible over the rational numbers. Hence, the factored form of the polynomial is:
(x^4 - 3x^3 - 11x^2 + 3x + 10) = (x - 1)(x + 1)(x^2 + x - 10)