Question

Plezzzzz tomorrow is my exam

Answer 1

hiii,

[12] x^2 -(b-2)x -2b
=x^2 - bx - 2x - 2b
=x (x - b) - 2 (x - b)
=(x - b) (x - 2)

[11] x^2 -(a -3)x -3a
= x^2 - ax - 3x - 3a
= x (x - a) - 3 (x - a)
=(x - a)(x - 3)

hope helps

Answer 2

see here i take common for (i)

3a²(b-4) - 9(b-4)=0
(3a²-9)(b-4)=0
a²=3 or b=4
a=±√3

(ii) we know that x²-(a+b)x+ab =0
has two solution a and b

then (ii) has two solution a and -3

(iii) from second has two solution b and -2