Question

What are the remaining linear factors of the polynomial 4x4−20x3+23x2+5x−6 with known factors (x−3) and (x−2) ?(1 point) Responses (x+3) and (x+2) open paren x plus 3 close paren and open paren x plus 2 close paren - no response given (2x−1) and (2x+1) open paren 2 x minus 1 close paren and open paren 2 x plus 1 close paren - no response given (4x4−20x3) and (23x2+5x−6) open paren 4 x to the 4th power minus 20 x cubed close paren and open paren 23 x squared plus 5 x minus 6 close paren - incorrect (4x2−1)

Answer 1

To determine the remaining linear factors of the polynomial \( 4x^4 - 20x^3 + 23x^2 + 5x - 6 \) given the known factors \( (x - 3) \) and \( (x - 2) \), we can use polynomial division or synthetic division to divide the polynomial by the factors \( (x - 3) \) and \( (x - 2) \).

### Step 1: Synthetic Division by \( (x - 3) \)

We start by dividing the polynomial by \( (x - 3) \):

1. Write down the coefficients:
\( 4, -20, 23, 5, -6 \).
2. Set up synthetic division. We will use \( 3 \) as the value.

\[
\begin{array}{r|rrrrr}
3 & 4 & -20 & 23 & 5 & -6 \\
& & 12 & -24 & -3 & 6 \\
\hline
& 4 & -8 & -1 & 2 & 0 \\
\end{array}
\]

The resulting polynomial is \( 4x^3 - 8x^2 - x + 2 \).

### Step 2: Synthetic Division by \( (x - 2) \)

Now we divide \( 4x^3 - 8x^2 - x + 2 \) by \( (x - 2) \):

1. Set up synthetic division again with the coefficients \( 4, -8, -1, 2 \) and the value \( 2 \).

\[
\begin{array}{r|rrrr}
2 & 4 & -8 & -1 & 2 \\
& & 8 & 0 & -2 \\
\hline
& 4 & 0 & -1 & 0 \\
\end{array}
\]

The resulting polynomial is \( 4x^2 - 1 \).

### Step 3: Factor the Remaining Polynomial

Now we have \( 4x^2 - 1 \). This can be factored further as a difference of squares:

\[
4x^2 - 1 = (2x - 1)(2x + 1)
\]

### Conclusion

So, the complete factorization of the given polynomial \( 4x^4 - 20x^3 + 23x^2 + 5x - 6 \) is:

\[
(x - 3)(x - 2)(2x - 1)(2x + 1)
\]

Thus, the remaining linear factors of the polynomial are \( (2x - 1) \) and \( (2x + 1) \).