Question

what must be added to 6x^5+5x^4+11x^3-3x^2+x+5 so that it may be exactly divisible by 3x^2-2x+4

Answer 1

Can you divide (6x^5 + 5x^4 + 11x^3 + 3x^2 + x + 5) by (3x^2 - 2x + 4) ?

Answer 2

yes I can

Answer 3

Good, then go on:

Divide and then tell me the remainder

Answer 4

the remainder is 15x^2-11x+5

Answer 5

Are you sure you did the division correct ?



Now, you have to add -ve of remainder to the dividend to make it completely divisible. So,

required expression to be added = -(-17x+17) = 17x - 17