Question

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1 Find the remainder when x⁹⁹⁹ is divided by x² - 4x + 3

2 Find the remainder when x⁵ is divided by x³ - 4x

3 Find the remainder when x³³ is divided by x² - 3x - 4

4 The volume of a cube with diagonal " d " will be

5 The remainder of a polynomial f(x) in x are 10 and 15 respectively when f(x) is divided by ( x -3 ) and ( x - 4 ) , Find the remainder when f(x) is divided by (x - 3)(x - 4)

6 When a third degree polynomial f(x) is divided by ( x - 3 ) , the quotient is Q(x) and the remainder is 0 . Also when f(x) is divided by [ Q(x) + x + 1 ], the quotient is ( x - 4 ) and the remainder is R(x) , Find the remainder R(x)

Answer 1

x²-4x+3

= x²-x-3x+3

=(x-1)(x-3)

Now Checking a small property; 19/2 gives remainder 1 ,and 19/7 gives remainder 5 . 19/2×7 = gives remainder 5 which is greatest remainder in both . 20/3 gives remainder 2 and 20/6 gives remainder 2 now 20/6*3 gives remainder 2 which is greatest .

From this, let f(x) be a polynomial then a, b are the remainders when divided by (x-e) , (x-f) respectively then for deg(x-e)(x-f) < deg f(x) then f(x) / (x-e) (x-f) yields a remainder either a if ( a > b)

Now (x-3) (x-1) which yields x=3, 1 Now
The question reduces to x^999/(x-3)(x-1) which will be done by remainder theorem .



Now the remainder of x^999 /( x-1) is 1 [ by remainder theorem]

Now x^999 / x-3 gives remainder 3^999 .

So, from the theory, x^999 when divided by (x-3)(x-1) gives a remainder 3^999 .

2) Now x³-4x=0

x(x²-4) = 0

x(x+2)(x-2) = 0

Now x = 0,2,-2 .

x^5 / x gives a remainder 0 .

x^5/(x+2) gives a remainder -32 .

x^5/x-2 gives a remainder 32 .

As 32>0>-32 .

x^5 / x³-4x gives a remainder 32 .

3) x²-3x-4 =0

x= -(-3)±√3²-4(1)(-4) / 2(1) = 3±√3²+4²/2

x= 3 ±5 / 2 = 4 or -1 .

Now( x+1)(x-4)

x³³/(x+1) gives a remainder -1
X³³ /(x-4) gives a remainder 4³³

Here -1<4³³

Hence, x³³/x²-3x-4 leaves a remainder 4³³

4) We know that in a cube ,diagonal = √3a²

Let diagonal be d =√3 a; a = d/√3 .

Volume of a cube = a³ = (d/√3)³ = d³/3√3

=√3d³/9

Volume of a cube with diagonal d is d²/3√3 or √3d³/9 .

5) 5) as we discussed earlier.
remainder of f(x) / x-3 = 10
remainders of f(x)/ x-4 = 15 .

As 15>10
Remainder of f(x) / (x-3)(x-4) = 15

6) From the question 6 ;
f(x) = q(x)*(x-3)

Now, f(x) - r(x) / [ q(x)+x+1] = x-4

q(x)*(x-3) - r(x) = (x-4)[q(x)+x+1]


q(x)*(x-3) - r(x) = [q(x)*x+x²+x] -4q(x)-4x-4 .

q(x)*x - 3q(x) - r(x) = q(x)*x + x²+x -4q(x)-4x-4 .

q(x)*x-3q(x)-q(x)*x -x² -x+4q(x)+4x+4 = r(x)

q(x) -x²+3x+4 = r(x) .


Hopefully it helps you!