Question

find the greatest number of 6 digit exactly divisible by 35,56,91 leaves remainder 7 in each case

Answer 1

maybe the question is wrong,

if it is correct  the do it like this

first remember that a= coefficent of x2 b= is the coefficient of your x term and c= is your non algebraic term.  

hence using your example a=1 b=3 and c=-4

In order to factorize this figure out which numbers can multiply to get the coefficient a. that would be 1x1=1

then say which 2 numbers can multiply to get c but also cross multiple with your a factors to give numbers that will add up to your b term.

the numbers that could do this is -1 and 4 because -1x4 =-4 but u can also say -1x1=-1 and 1x4=4 and add them up you get +3 which is your b term.  

then take your first factor of a and your first factor of c which would be 1 and -1 the first factor of a would become x so you now have (x-1)  

then take second factor of a and second factor of c and you have (x+4)  

remember your a factors become your Xs  

answer is therefore (x-1) (x+4)

Answer 2

greatest 6 digit number = 999999

LCM of 35, 56, 91 = 3640

999999÷ 3640 = 274 as quotient and 2639 as reminder

now 999999 - 2639  or 3640 x 274 = 997360 is the greatest number which is divisible by 35, 56, and 91

Now add 7 to this number 997360 + 7 = 997367 is the greatest number which will have 7 as reminder