To find the additional linear factors of the polynomial \( P(x) = x^3 - 6x^2 + 11x - 5 \) given that \( x - 3 \) is a factor, we can perform polynomial long division or synthetic division to divide \( P(x) \) by \( x - 3 \).
Synthetic Division
Using synthetic division with \( x = 3 \):
- Write down the coefficients of \( P(x) \): \( 1, -6, 11, -5 \).
- Set up synthetic division:
3 | 1 -6 11 -5
| 3 -9 6
---------------------
1 -3 2 1
- Calculate:
- Bring down the 1.
- Multiply 3 (the root) by 1 (the first coefficient), which gives 3. Add this to -6 to get -3.
- Multiply 3 by -3 to get -9. Add this to 11 to get 2.
- Multiply 3 by 2 to get 6. Add this to -5 to get 1.
The result of the synthetic division gives us a quotient of:
\[ Q(x) = x^2 - 3x + 2 \]
And a remainder of \( 0 \) (which confirms that \( x - 3 \) is indeed a factor).
Factoring the Quotient
Now, we can factor \( Q(x) = x^2 - 3x + 2 \):
We need to find two numbers that multiply to \( 2 \) (the constant term) and add to \( -3 \) (the linear coefficient). These two numbers are \( -1 \) and \( -2 \).
Thus, we can factor \( Q(x) \) as:
\[ Q(x) = (x - 1)(x - 2) \]
Conclusion
Putting this all together, the polynomial \( P(x) \) can be fully factored as:
\[ P(x) = (x - 3)(x - 1)(x - 2) \]
Therefore, the additional linear factors of \( P(x) \) besides \( x - 3 \) are \( x - 1 \) and \( x - 2 \).
1 answer