Question

What must be added to 2x²-5x +6 to get x³-3x²+3x-5

Answer 1

let T must be added

2x²-5x+6+T = x³-3x²+3x-5

T= x³-3x²-2x²+3x+5x-5-6

T= x³-5x²+8x-11

Answer 2

Let the tens digit of the required number be x and the units digit be y. Then,

x+y=12 .........(1)

Required Number = (10x+y).

Number obtained on reversing the digits = (10y+x).

Therefore,

(10y+x)−(10x+y)=18

9y−9x=18

y−x=2 ..........(2)

On adding (1) and (2), we get,

2y=14⟹y=7

Therefore,

x=5

Hence, the required number is 57.