Answer :
To factor the polynomial \( 4x^3 + 5x^2 - 18x + 9 \) and identify which of the given factors are correct, let's use synthetic division for each candidate factor. We'll check if each factor, when used, leaves a remainder of zero, which would confirm that it is indeed a factor of the polynomial.
### Step-by-Step Factor Verification
#### Synthetic Division for \( x-1 \)
1. Set \( x \) equal to \( 1 \):
[tex]\[ 1 \][/tex]
2. Coefficients of the polynomial \( 4x^3 + 5x^2 - 18x + 9 \):
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 1 & 4 & 5 & -18 & 9 \\ & & 4 & 9 & -9 \\ \hline & 4 & 9 & -9 & 0 \\ \end{array} \][/tex]
Since the remainder is \( 0 \), \( x-1 \) is a factor.
#### Synthetic Division for \( 2x-3 \)
1. Set \( x = \frac{3}{2} \):
[tex]\[ 1.5 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 1.5 & 4 & 5 & -18 & 9 \\ & & 6 & 16.5 & -0.75 \\ \hline & 4 & 11 & -1.5 & 8.25 \\ \end{array} \][/tex]
Since the remainder is \( 8.25 \), \( 2x-3 \) is not a factor.
#### Synthetic Division for \( x-3 \)
1. Set \( x \) equal to \( 3 \):
[tex]\[ 3 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 3 & 4 & 5 & -18 & 9 \\ & & 12 & 51 & 99 \\ \hline & 4 & 17 & 33 & 108 \\ \end{array} \][/tex]
Since the remainder is \( 108 \), \( x-3 \) is not a factor.
#### Synthetic Division for \( x+3 \)
1. Set \( x \) equal to \( -3 \):
[tex]\[ -3 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} -3 & 4 & 5 & -18 & 9 \\ & & -12 & 21 & -9 \\ \hline & 4 & -7 & 3 & 0 \\ \end{array} \][/tex]
Since the remainder is \( 0 \), \( x+3 \) is a factor.
#### Synthetic Division for \( 4x-3 \)
1. Set \( x = \frac{3}{4} \):
[tex]\[ 0.75 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 0.75 & 4 & 5 & -18 & 9 \\ & & 3 & 6 & -9 \\ \hline & 4 & 8 & -12 & 0 \\ \end{array} \][/tex]
Since the remainder is \( 0 \), \( 4x-3 \) is a factor.
#### Synthetic Division for \( 2x+1 \)
1. Set \( x = -\frac{1}{2} \):
[tex]\[ -0.5 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} -0.5 & 4 & 5 & -18 & 9 \\ & & -2 & 1 & -0.5 \\ \hline & 4 & 3 & -17 & 8.5 \\ \end{array} \][/tex]
Since the remainder is \( 8.5 \), \( 2x+1 \) is not a factor.
### Conclusion
Therefore, the correct factors of the polynomial \( 4x^3 + 5x^2 - 18x + 9 \) are:
[tex]\[ x-1, \; x+3, \; 4x-3 \][/tex]
The other given factors do not result in a zero remainder when used in synthetic division with the polynomial.
### Step-by-Step Factor Verification
#### Synthetic Division for \( x-1 \)
1. Set \( x \) equal to \( 1 \):
[tex]\[ 1 \][/tex]
2. Coefficients of the polynomial \( 4x^3 + 5x^2 - 18x + 9 \):
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 1 & 4 & 5 & -18 & 9 \\ & & 4 & 9 & -9 \\ \hline & 4 & 9 & -9 & 0 \\ \end{array} \][/tex]
Since the remainder is \( 0 \), \( x-1 \) is a factor.
#### Synthetic Division for \( 2x-3 \)
1. Set \( x = \frac{3}{2} \):
[tex]\[ 1.5 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 1.5 & 4 & 5 & -18 & 9 \\ & & 6 & 16.5 & -0.75 \\ \hline & 4 & 11 & -1.5 & 8.25 \\ \end{array} \][/tex]
Since the remainder is \( 8.25 \), \( 2x-3 \) is not a factor.
#### Synthetic Division for \( x-3 \)
1. Set \( x \) equal to \( 3 \):
[tex]\[ 3 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 3 & 4 & 5 & -18 & 9 \\ & & 12 & 51 & 99 \\ \hline & 4 & 17 & 33 & 108 \\ \end{array} \][/tex]
Since the remainder is \( 108 \), \( x-3 \) is not a factor.
#### Synthetic Division for \( x+3 \)
1. Set \( x \) equal to \( -3 \):
[tex]\[ -3 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} -3 & 4 & 5 & -18 & 9 \\ & & -12 & 21 & -9 \\ \hline & 4 & -7 & 3 & 0 \\ \end{array} \][/tex]
Since the remainder is \( 0 \), \( x+3 \) is a factor.
#### Synthetic Division for \( 4x-3 \)
1. Set \( x = \frac{3}{4} \):
[tex]\[ 0.75 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} 0.75 & 4 & 5 & -18 & 9 \\ & & 3 & 6 & -9 \\ \hline & 4 & 8 & -12 & 0 \\ \end{array} \][/tex]
Since the remainder is \( 0 \), \( 4x-3 \) is a factor.
#### Synthetic Division for \( 2x+1 \)
1. Set \( x = -\frac{1}{2} \):
[tex]\[ -0.5 \][/tex]
2. Coefficients:
[tex]\[ 4, 5, -18, 9 \][/tex]
3. Perform the synthetic division:
[tex]\[ \begin{array}{r|rrrr} -0.5 & 4 & 5 & -18 & 9 \\ & & -2 & 1 & -0.5 \\ \hline & 4 & 3 & -17 & 8.5 \\ \end{array} \][/tex]
Since the remainder is \( 8.5 \), \( 2x+1 \) is not a factor.
### Conclusion
Therefore, the correct factors of the polynomial \( 4x^3 + 5x^2 - 18x + 9 \) are:
[tex]\[ x-1, \; x+3, \; 4x-3 \][/tex]
The other given factors do not result in a zero remainder when used in synthetic division with the polynomial.
