Who can solve this difficult question....10 points...?

Question

Given that 2x^2 + 3px - 2q and x^2 + q have common factor x-a, where p,q and a are non zero constants.Show that 9p^2 + 16q =0

Answer 1

x^2 + q has common factor x-a
so, x^2 +q = (x-a) (x-b)
x^2 + q = x^2 - (a+b)x + ab
a+ b = 0
b = -a

q= ab
q = a (-a)
q = -a^2

2x^2 +3 px -2q has common factor x-a
Divide 2x^2 +3 px -2 by 2
so x^2 +3p/2x-q = (x-a)(x-c)
x^2 +3p/2x -q = x^2 -(a+c)x +ac

-q = ac
q = -ac
-a^2 = -ac
c=a

3p/2 = -(a+c)
3p/2 = -(a+a)
p = -4/3a

Thus,
9p^2 + 16q
= 9 (-4/3a)^2 + 16(-a^2)
= 9 (16/9)a^2 -16a^2
= 0!!!

Answer 2

Quite simple!

I am using the long polynomial division method.
Its difficult to show the work out though!

Divide 2x^2 + 3px - 2q by x-a
Quotient = 2x + 3p+ 2a
Remainder = -2q + 3ap+ 2a^2
Since x-a is a factor, remainder = 0
Thus 2q = 3ap+ 2a^2....(1)

Divide x^2 + q by x-a
Quotient = x+a
Remainder = q+a^2
Since x-a is a factor, remainder = 0
Thus q + a^2 = 0....(2)

Substitute value of a^2 from equation (2) in equation (1)
we get:
2q = 3ap -2q
4q = 3ap

Squaring both sides
16q^2 = 9*a^2*p^2

Substitute value of a^2 from equation (2) in equation (1)
16q^2 = 9*(-q)*p^2

Thus
9p^2 + 16q =0

Answer 3

I hate Algebra, or whatever the heck that is.