what is the answer?

Question

1) For the quadratic equation, x^2 - 4x + c = 0 where the equation has only one solution... Find the value of c.

How do I do that? Full working out please! The answer is 4 but I dont know why!!!!! 

2) I also suck at factorising. I can factorise using bridge method, cross method, etc....but the really hard ones...... like this one is hard!!! 

What are the steps required to solve this??? 
TRIAL AND ERROR? OR AN ACTUAL METHOD??? 

Factorise: 16x^2 - y^2 - 6y - 9 

The answer is: (4x - 3 -y) (4x + 3 + y) 

BUT I DONT KNOW WHY :( I used the cross method and stuff, but It still does work 

Factorise: a^2 - b^2 + a - b 

The answer is (a-b) (a+b+1) 

I get that when you factorise it, it becomes: (a-b)(a+b)+(a-b) 
but HOW does it change into (a-b)(a+b+1) 

Answer 1

1) The equation x² - 4x + c = 0

Compare it to quadratic equation ax² + bx + c = 0

a = 1, b = - 4, c = c

The equation has only one solution.

So, discriminant b² - 4ac = 0

(- 4)² - 4(1)(c) = 0

16 - 4c = 0

4c = 16

c = 16/4

Slution c = 4.

Answer 2

2) The expression 16x² - y² - 6y - 9

= 16x² - (y² + 6y + 9)

= [(4x)² - (y² + 2(3)y + 3²]

{ Apply the formula (a + b)² = a² + b² + 2ab, in this case a = y, b = 3 }

= (4x)² - (y + 3)²

{ Apply the formula a² - b² = (a + b)(a - b), in this case a = 4x, b = y + 3 }

= [ 4x + (y + 3)][4x - (y + 3)]

Factoring of 16x² - y² - 6y - 9 = (4x + y + 3)(4x - y - 3).

Answer 3

3) The expression a² - b² + a - b

Group the terms into two pairs.

= (a² - b²) + (a - b)

{ Apply the formula a² - b² = (a + b)(a - b) }

= [(a + b)(a - b)] + (a - b)

Take out common term ( a - b) .

= (a - b) [ (a + b) + 1]

= (a - b) (a + b + 1)

Factoring of a² - b² + a - b = (a - b)(a + b + 1)