Using the Remainder Theorem, factorise each of the following completely:
(i) `3x^3+2x^(2)-19x+6`
(ii) `2x^(3)+x^2-13x+6`
(iii) `3x^(3)+2x^2-23x-30`
(iv) `4x^3+7x^2-36x-63`
(v) `x^3+x^2-4x-4`

Answer

To factorise the given polynomials using the Remainder Theorem, we will follow these steps for each part:

### (i) Factorise `3x^3 + 2x^2 - 19x + 6`

1. **Find a root using the Remainder Theorem**: We will test possible rational roots. Let's try \( x = 2 \):
\[ 3(2)^3 + 2(2)^2 - 19(2) + 6 = 3(8) + 2(4) - 38 + 6 = 24 + 8 - 38 + 6 = 0 \]
Thus, \( x = 2 \) is a root, so \( x - 2 \) is a factor.

2. **Perform synthetic division** of \( 3x^3 + 2x^2 - 19x + 6 \) by \( x - 2 \):
\[ \begin{array}{r|rrrr} 2 & 3 & 2 & -19 & 6 \\ & & 6 & 16 & -6 \\ \hline & 3 & 8 & -3 & 0 \\ \end{array} \]
The result is \( 3x^2 + 8x - 3 \).

3. **Factor the quadratic** \( 3x^2 + 8x - 3 \):
We can factor this as:
\[ 3x^2 + 9x - x - 3 = 3x(x + 3) - 1(x + 3) = (3x - 1)(x + 3) \]

4. **Final factorization**:
\[ 3x^3 + 2x^2 - 19x + 6 = (x - 2)(3x - 1)(x + 3) \]

### (ii) Factorise `2x^3 + x^2 - 13x + 6`

1. **Find a root**: Testing \( x = 2 \):
\[ 2(2)^3 + (2)^2 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0 \]
Thus, \( x - 2 \) is a factor.

2. **Synthetic division**:
\[ \begin{array}{r|rrrr} 2 & 2 & 1 & -13 & 6 \\ & & 4 & 10 & -6 \\ \hline & 2 & 5 & -3 & 0 \\ \end{array} \]
The result is \( 2x^2 + 5x - 3 \).

3. **Factor the quadratic**:
\[ 2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3) \]

4. **Final factorization**:
\[ 2x^3 + x^2 - 13x + 6 = (x - 2)(2x - 1)(x + 3) \]

### (iii) Factorise `3x^3 + 2x^2 - 23x - 30`

1. **Find a root**: Testing \( x = -2 \):
\[ 3(-2)^3 + 2(-2)^2 - 23(-2) - 30 = -24 + 8 + 46 - 30 = 0 \]
Thus, \( x + 2 \) is a factor.

2. **Synthetic division**:
\[ \begin{array}{r|rrrr} -2 & 3 & 2 & -23 & -30 \\ & & -6 & 8 & 30 \\ \hline & 3 & -4 & -15 & 0 \\ \end{array} \]
The result is \( 3x^2 - 4x - 15 \).

3. **Factor the quadratic**:
\[ 3x^2 - 9x + 5x - 15 = 3x(x - 3) + 5(x - 3) = (3x + 5)(x - 3) \]

4. **Final factorization**:
\[ 3x^3 + 2x^2 - 23x - 30 = (x + 2)(3x + 5)(x - 3) \]

### (iv) Factorise `4x^3 + 7x^2 - 36x - 63`

1. **Find a root**: Testing \( x = 3 \):
\[ 4(3)^3 + 7(3)^2 - 36(3) - 63 = 108 + 63 - 108 - 63 = 0 \]
Thus, \( x - 3 \) is a factor.

2. **Synthetic division**:
\[ \begin{array}{r|rrrr} 3 & 4 & 7 & -36 & -63 \\ & & 12 & 57 & 63 \\ \hline & 4 & 19 & 21 & 0 \\ \end{array} \]
The result is \( 4x^2 + 19x + 21 \).

3. **Factor the quadratic**:
\[ 4x^2 + 12x + 7x + 21 = 4x(x + 3) + 7(x + 3) = (4x + 7)(x + 3) \]

4. **Final factorization**:
\[ 4x^3 + 7x^2 - 36x - 63 = (x - 3)(4x + 7)(x + 3) \]

### (v) Factorise `x^3 + x^2 - 4x - 4`

1. **Find a root**: Testing \( x = 2 \):
\[ (2)^3 + (2)^2 - 4(2) - 4 = 8 + 4 - 8 - 4 = 0 \]
Thus, \( x - 2 \) is a factor.

2. **Synthetic division**:
\[ \begin{array}{r|rrrr} 2 & 1 & 1 & -4 & -4 \\ & & 2 & 6 & 4 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array} \]
The result is \( x^2 + 3x + 2 \).

3. **Factor the quadratic**:
\[ x^2 + 2x + x + 2 = (x + 2)(x + 1) \]

4. **Final factorization**:
\[ x^3 + x^2 - 4x - 4 = (x - 2)(x + 2)(x + 1) \]