Help, Math problem?

Question

Factorise:

16a^2 - 25b^2c^2

Also Solve these simultaneous equations:

x-y=2

y=2(x^2 - x - 2)

Answer 1

16a^2 - 25b^2c^2
=(4a - 5bc)(4a + 5bc)

second part:
rearrange first equation y = x - 2
sub that into first equation and distribute the two
x - 2 = 2x^2 - 2x - 4
0 = 2x^2 - 3x -2
0 = (2x + 1)(x - 2)
x = -1/2 or x = 2

Answer 2

16a^2 - 25b^2c^2
is a difference-of-squares.
(a^2 - b^2) = (a + b)(a - b)

x - y = 2
y = 2(x^2 - x - 2)
by substitution,
x - 2 = 2(x^2 - x - 2)
2x^2 - 2x - 4 = x - 2
2x^2 - 3x - 2 = 0
(2x + 1)(x - 2) = 0
x = - 1/2, 2

Answer 3

16a^2 - 25b^2c^2 = (4a - 5bc)(4a + 5bc)

--------------------------

x - y = 2
y = 2(x^2 - x - 2)

x - 2(x^2 - x - 2) = 2
x - 2x^2 + 2x + 4 = 2
-2x^2 + 3x + 2 = 0
2x^2 - 3x - 2 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-3) ± sqrt((-3)^2 - 4(2)(-2)))/(2(2))
x = (3 ± sqrt(9 + 16))/4
x = (3 ± sqrt(25))/4
x = (3 ± 5)/4
x = (8/4) or (-2/4)
x = 2 or (-1/2)

2 - y = 2
-y = 0
y = 0
or
(-1/2) - y = 2
-y = (5/2)
y = (-5/2)

ANS : (2,0) or ((-1/2),(-5/2))