2.a) x^2+6x+c
All the integers are positive in this equation.
Quadratic a are factorized to give (x+r)(x+s). As all integers are positive, in this problem r and s are positive.
C is equal to r multiplied by s, and r and s add up to give the coefficient of x (in this case 6).
To find c we must find all possible pairs of integers that give the sum of 6.
These are: 1+5, 2+4, 3+3
We then multiply them to give:
C= 1 x 5 = 5
2 x 4 = 8
3 x 3 = 9
2.b) x^2-10x+c
This time the coefficient of x is negative (-10). However c is positive, so both r and s must be negative, so they multiply to give a positive number.
Once again r and s must multiply to give c, and r and s must add to give -10.
The pairs of integers that are negative and equal -10 are: -9+ -1, -8+ -2, -7+ -3, -6+ -4 and -5+ -5
If we multiply each pair to give c, we get
C= -9 x-1 = 9
-8 x-2 = 16
-7 x-3 = 21
-6 x-4 = 24
-5 x-5 = 25