How, let a = a rational number. let b = an irrational number. assume a + b = c and c is rational (attempting to disprove the conjecture). a + b = c b = c - a explain how alfred's argument contradicts his initial assumption proving that the sum cannot be rational?

If 'a' is a rational number and c is rational, then 
a = p/q
c = r/s
where p,q,r,s are integers (q and s can't be zero)

Subtracting c-a gives
b = c-a
b = (p/q) - (r/s)
b = (ps/qs) - (qr/qs)
b = (ps - qr)/(qs)

The quantity pq - qr is an integer. The reason why is because ps and qr are both integers (multiplying any two integers leads to another integer). Subtracting any two integers results in another integer.

So we have (ps - qr)/(qs) in the form (integer)/(integer) = rational number

Therefore, b is a rational number, but this contradicts the given info that b is irrational. If b is irrational, then we CANNOT write it as a ratio of integers.

This contradiction proves the assumption "a+b = c and c is rational" is incorrect
The sum is irrational.

Therefore, if a+b = c, where 'a' is rational and b is irrational, then c is irrational.

p(x) × q(x)

= (3x² - x + 1) × (2x² + 3x - 2)

= 6x⁴ + 9x³ - 6x² - 2x³ - 3x² + 2x + 2x² + 3x - 2

= 6x⁴ + 7x³ - 7x² + 5x - 2