Question

Find the remainder when x 33 is divided by x^2-3x-4

Answer 1

Answer:

The remainder is ($\bold{4^{33}}$+1)x/5+$\bold{4^{33}}$-4/5

Step-by-step explanation:

Step 1:

Here Given Data: $x^{33}$ is divided by $x^{2}$-3 x-4

To calculate the Remainder:

The value of  $x^{2}$-3 x-4= (x+1)(x-4)

$x^{33}$ / $x^{2}$-3 x-4= $x^{33}$ /(x+1)(x-4)

Step 2:

Now,

$x^{33}$=(x+1)(x-4)q(x)+ax+b  (where ax+b is remainder and q(x) is quotient)

At  x=-1

Step 3:

-a+b=-1.....(i)

At  x=4

4a+b= $4^{33}$...(ii)

Step 4:

From equation (i) and equation (ii) we get :-

a=$4^{33}$+1/5 and then putting this value to the equation we get b=4^{33}-4/5

Step 5:

Therefore,  Remainder =ax+b

=($4^{33}$+1)x/5+$4^{33}$-4/5

Answer 2

this is the answer to this question