Find the remainder when `x^3+3x^2+3x+1` is divided by (i) `x+1` (ii) `x-1/2` (iii) `x` (iv) `x+pi` (v) `5+2x`

Explanation

"Solution:
Let `p(x)` be any polynomial of degree greater than or equal to one and let 'a' be any real number.
If a polynomial `p(x)` is divided by `x - a` then the remainder is p(a).
Let `p(x) = x³ + 3x² + 3x + 1`
(i) The root of `x + 1 = 0` is `-1`
`p(-1)=(-1)3 +3(-1)2 + 3(-1) +1)-13-3+1= 0`
Hence by the remainder theorem, `0` is the remainder when `x³ + 3x² + 3x+1`is divided by `x + 1`. We can also say that `x + 1` is a factor of `x³ + 3x² + 3x+1`
(ii) The root of `x- (1/2) = 0` is `1/2`
`p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1`
= `1/8+3/4+3/2+1`
= `(1+6+12+8)/8 = 27/8`
Hence by the remainder theorem, `27/8` is the remainder when `x³ + 3x² +3x + 1` is divided by `x`

(iii) The root of `x = 0` is `0`
`p(0) = (0)³ +3(0)2 + 3(0) +1
= 0+0+0+1=1`
Hence by the remainder theorem, `1` is the remainder when `x³ + 3x² + 3x +1`
is divided by `x`
(iv) The root of `x + π = 0` is `-π`
`р (-π) = (-1) ³ + 3 (-π) ² + 3 (-π) + 1`
Hence by the remainder theorem, `-π³ +3π² - 3π+1` is the remainder
when `x³ + 3x² + 3x +1` is divided by `x + π`.
(v) Now, the root of `5 + 2x = 0` is `-5/2`
`p(-5/2) = [(-5/2)³ + 3(-5/2)2 + 3(-5/2) + 1]`
= `[(-125/8) + (75/4) + (-15/2) + 1]`
= `(-125+150-60+8)/8`
= `(-185+158) / 8`
= `-27/8`
Hence by remainder theorem, `-27/8` is the remainder when `x³ + 3x² + 3x+ 1` is divided by `5 + 2x`."