Using remainder theorem, find the remainder when :
`(i) x^(3)+5x^(2)-3` is divided by `(x-1) " " (ii) x^(4)-3x^(2)+2` is divided by (x-2)
`(iii)2x^(3)+3x^(2)-5x+2` is divided by `(x+3) " " (iv) x^(3)+2x^(2)-x+1` is divided by (x+2)
`(v) x^(3)+3x^(2)-5x+4` is divided by `(2x-1) " " (vi) 3x^(3)+6x^(2)-15x+2` is divided by (3x-1)

Answer

To find the remainder when a polynomial is divided by a linear polynomial using the Remainder Theorem, we can follow these steps:

### Solution Steps:

1. **Identify the Polynomial and the Divisor**:
- For each part, identify the polynomial \( f(x) \) and the divisor \( x - a \).

2. **Use the Remainder Theorem**:
- According to the Remainder Theorem, the remainder of \( f(x) \) when divided by \( x - a \) is simply \( f(a) \).

3. **Calculate \( f(a) \)**:
- Substitute \( a \) into the polynomial \( f(x) \) to find the remainder.

### Detailed Solutions:

**(i)** For \( f(x) = x^3 + 5x^2 - 3 \) and divisor \( x - 1 \):
- Here, \( a = 1 \).
- Calculate \( f(1) = 1^3 + 5(1^2) - 3 = 1 + 5 - 3 = 3 \).
- **Remainder**: 3

**(ii)** For \( f(x) = x^4 - 3x^2 + 2 \) and divisor \( x - 2 \):
- Here, \( a = 2 \).
- Calculate \( f(2) = 2^4 - 3(2^2) + 2 = 16 - 12 + 2 = 6 \).
- **Remainder**: 6

**(iii)** For \( f(x) = 2x^3 + 3x^2 - 5x + 2 \) and divisor \( x + 3 \):
- Here, \( a = -3 \).
- Calculate \( f(-3) = 2(-3)^3 + 3(-3)^2 - 5(-3) + 2 = -54 + 27 + 15 + 2 = -10 \).
- **Remainder**: -10

**(iv)** For \( f(x) = x^3 + 2x^2 - x + 1 \) and divisor \( x + 2 \):
- Here, \( a = -2 \).
- Calculate \( f(-2) = (-2)^3 + 2(-2)^2 - (-2) + 1 = -8 + 8 + 2 + 1 = 3 \).
- **Remainder**: 3

**(v)** For \( f(x) = x^3 + 3x^2 - 5x + 4 \) and divisor \( 2x - 1 \):
- Here, \( a = \frac{1}{2} \).
- Calculate \( f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 5\left(\frac{1}{2}\right) + 4 = \frac{1}{8} + \frac{3}{4} - \frac{5}{2} + 4 \).
- Convert to a common denominator (8):
- \( \frac{1}{8} + \frac{6}{8} - \frac{20}{8} + \frac{32}{8} = \frac{1 + 6 - 20 + 32}{8} = \frac{19}{8} \).
- **Remainder**: \( \frac{19}{8} \)

**(vi)** For \( f(x) = 3x^3 + 6x^2 - 15x + 2 \) and divisor \( 3x - 1 \):
- Here, \( a = \frac{1}{3} \).
- Calculate \( f\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^3 + 6\left(\frac{1}{3}\right)^2 - 15\left(\frac{1}{3}\right) + 2 \).
- Simplifying gives:
- \( 3 \cdot \frac{1}{27} + 6 \cdot \frac{1}{9} - 5 + 2 = \frac{1}{9} + \frac{2}{3} - 5 + 2 \).
- Convert to a common denominator (9):
- \( \frac{1}{9} + \frac{6}{9} - \frac{45}{9} + \frac{18}{9} = \frac{1 + 6 - 45 + 18}{9} = \frac{-20}{9} \).
- **Remainder**: \( -\frac{20}{9} \)