Using remainder theorem, factorise : ` x^(3) + 7x^(2) - 21 x - 27 ` Completely and then solve ` x^(3) + 7x^(2) - 21 x - 27=0 `

Answer

To factorize the polynomial \( f(x) = x^3 + 7x^2 - 21x - 27 \) using the Remainder Theorem and then solve the equation \( f(x) = 0 \), follow these steps:

### Step 1: Identify the polynomial
Let \( f(x) = x^3 + 7x^2 - 21x - 27 \).

### Step 2: Use the Remainder Theorem
We will check for possible rational roots using the Rational Root Theorem. We will start by testing \( x = -1 \).

### Step 3: Evaluate \( f(-1) \)
Calculate:
\[ f(-1) = (-1)^3 + 7(-1)^2 - 21(-1) - 27 \]
\[ = -1 + 7 + 21 - 27 \]
\[ = -1 + 7 + 21 - 27 = 0 \]

Since \( f(-1) = 0 \), \( x + 1 \) is a factor of \( f(x) \).

### Step 4: Perform polynomial long division
Now, we will divide \( f(x) \) by \( x + 1 \).

1. Divide \( x^3 \) by \( x \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x + 1 \) to get \( x^3 + x^2 \).
3. Subtract \( (x^3 + x^2) \) from \( f(x) \):
\[ (x^3 + 7x^2 - 21x - 27) - (x^3 + x^2) = 6x^2 - 21x - 27 \]

4. Now divide \( 6x^2 \) by \( x \) to get \( 6x \).
5. Multiply \( 6x \) by \( x + 1 \) to get \( 6x^2 + 6x \).
6. Subtract \( (6x^2 + 6x) \):
\[ (6x^2 - 21x - 27) - (6x^2 + 6x) = -27x - 27 \]

7. Divide \( -27x \) by \( x \) to get \( -27 \).
8. Multiply \( -27 \) by \( x + 1 \) to get \( -27x - 27 \).
9. Subtract:
\[ (-27x - 27) - (-27x - 27) = 0 \]

Thus, we have:
\[ f(x) = (x + 1)(x^2 + 6x - 27) \]

### Step 5: Factor the quadratic \( x^2 + 6x - 27 \)
Now we will factor \( x^2 + 6x - 27 \) using the middle-term splitting method.

1. We need two numbers that multiply to \( -27 \) and add to \( 6 \). These numbers are \( 9 \) and \( -3 \).
2. Rewrite the quadratic:
\[ x^2 + 9x - 3x - 27 = (x^2 + 9x) + (-3x - 27) \]
3. Factor by grouping:
\[ x(x + 9) - 3(x + 9) = (x - 3)(x + 9) \]

### Step 6: Write the complete factorization
Thus, we can write:
\[ f(x) = (x + 1)(x - 3)(x + 9) \]

### Step 7: Solve \( f(x) = 0 \)
Set each factor to zero:
1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( x - 3 = 0 \) → \( x = 3 \)
3. \( x + 9 = 0 \) → \( x = -9 \)

### Final Solutions
The solutions to the equation \( x^3 + 7x^2 - 21x - 27 = 0 \) are:
\[ x = -1, \quad x = 3, \quad x = -9 \]