What are the additional linear factors x^3-6x^2+11x-5 if x-3 is a factor

1 answer

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 \( x = 3 \):

  1. Write down the coefficients of \( P(x) \): \( 1, -6, 11, -5 \).
  2. Set up synthetic division:
3 |  1   -6   11   -5
   |       3   -9    6
   ---------------------
     1   -3    2    1
  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 \( x - 2 \).