(3x + 5) is a factor of the polynomial `(a-1)x^(3)+(a+1)x^2-(2a+1)x-15`. Find the value of 'a'. For this value of a'. Factorise the given polynomial completely

Answer

To solve the problem, we need to find the value of \( a \) such that \( (3x + 5) \) is a factor of the polynomial

\[ P(x) = (a - 1)x^3 + (a + 1)x^2 - (2a + 1)x - 15. \]

### Step 1: Set up the equation
Since \( (3x + 5) \) is a factor, we can set \( 3x + 5 = 0 \) to find the value of \( x \) at which \( P(x) \) should equal zero.

\[ 3x + 5 = 0 \implies x = -\frac{5}{3}. \]

### Step 2: Substitute \( x = -\frac{5}{3} \) into \( P(x) \)
Now we substitute \( x = -\frac{5}{3} \) into the polynomial \( P(x) \):

\[ P\left(-\frac{5}{3}\right) = (a - 1)\left(-\frac{5}{3}\right)^3 + (a + 1)\left(-\frac{5}{3}\right)^2 - (2a + 1)\left(-\frac{5}{3}\right) - 15. \]

### Step 3: Calculate each term
Calculating each term separately:

1. \( \left(-\frac{5}{3}\right)^3 = -\frac{125}{27} \) so \( (a - 1)\left(-\frac{125}{27}\right) = -\frac{125(a - 1)}{27} \).
2. \( \left(-\frac{5}{3}\right)^2 = \frac{25}{9} \) so \( (a + 1)\left(\frac{25}{9}\right) = \frac{25(a + 1)}{9} \).
3. \( - (2a + 1)\left(-\frac{5}{3}\right) = \frac{5(2a + 1)}{3} \).

Putting it all together, we have:

\[ P\left(-\frac{5}{3}\right) = -\frac{125(a - 1)}{27} + \frac{25(a + 1)}{9} + \frac{5(2a + 1)}{3} - 15. \]

### Step 4: Simplify the equation
To simplify, we can multiply the entire equation by 27 to eliminate the denominators:

\[ -125(a - 1) + 75(a + 1) + 45(2a + 1) - 405 = 0. \]

Expanding this gives:

\[ -125a + 125 + 75a + 75 + 90a + 45 - 405 = 0. \]

Combining like terms:

\[ (-125a + 75a + 90a) + (125 + 75 + 45 - 405) = 0, \]
\[ 40a - 160 = 0. \]

### Step 5: Solve for \( a \)
Now, we can solve for \( a \):

\[ 40a = 160 \implies a = \frac{160}{40} = 4. \]

### Step 6: Factor the polynomial completely
Now that we have \( a = 4 \), we substitute back into the polynomial:

\[ P(x) = (4 - 1)x^3 + (4 + 1)x^2 - (2 \cdot 4 + 1)x - 15, \]
\[ P(x) = 3x^3 + 5x^2 - 9x - 15. \]

Since \( (3x + 5) \) is a factor, we can perform polynomial long division or synthetic division to factor \( P(x) \).

### Step 7: Perform the division
Dividing \( P(x) \) by \( 3x + 5 \):

1. Divide \( 3x^3 \) by \( 3x \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( 3x + 5 \) to get \( 3x^3 + 5x^2 \).
3. Subtract to get \( -9x - 15 \).
4. Divide \( -9x \) by \( 3x \) to get \( -3 \).
5. Multiply \( -3 \) by \( 3x + 5 \) to get \( -9x - 15 \).
6. Subtract to get \( 0 \).

Thus, we have:

\[ P(x) = (3x + 5)(x^2 - 3). \]

### Final Factorization
The complete factorization of the polynomial is:

\[ P(x) = (3x + 5)(x + \sqrt{3})(x - \sqrt{3}). \]

### Summary
The value of \( a \) is \( 4 \), and the complete factorization of the polynomial is:

\[ P(x) = (3x + 5)(x + \sqrt{3})(x - \sqrt{3}). \]