WHAT TO DO: 
HOW TO DO IT: 
1. Solve the equation x^{ 2} + 4(x+1)^{ 2} = (2x + 3)^{ 2}
Remove parentheses and simplify both sides:
Rewrite in standard equation form
and solve the equation: ⇒
Scratch: Find a numbers whose product is 5
with difference of 4 , larger sign "−".
Set each factor equal to zero:
The solution:

x^{ 2} + 4(x+1)^{ 2} = (2x + 3)^{ 2}
x^{ 2} + 4(x^{ 2}+2x+1) = 4x^{ 2} + 12x +9
x^{ 2} + 4x^{ 2}+8x+4 = 4x^{ 2} + 12x +9
x^{ 2} − 4x − 5 = 0
{5,1}
(x − 5)(x + 1) = 0
(x − 5) = 0 or (x + 1) = 0
x = 5 or x = − 1 
Check the original, equation
x^{ 2} + 4(x+1)^{ 2} = (2x + 3)^{ 2} 
⇒ x = 5
⇒ x = 1 
(5)^{2} + 4(5+1)^{2} = (2Â·5+3)^{2} , 25 + 144 =
139
(1)^{2} + 4(1+1)^{2} = (2Â·1+3)^{2} , 1 + 0 =
1 
2. Solve the equation x^{ 3} = 9x^{ 2} − 20x
Rewrite in standard equation form
and solve the equation: ⇒
Factor out the common factor x ⇒
Scratch: Solve the inner equation: ⇒
Find a pair of numbers whose
product is 20 with sum of 9. sign "−".
Set each factor equal to zero:
The solution:

x^{ 3} = 9x^{ 2} − 20x
x^{ 3} − 9x^{ 2} + 20x = 0
x (x^{ 2} − 9x + 20) = 0
x (x^{ 2} − 9x + 20) = 0
x (x − 5)(x − 4) = 0
x = 0, (x − 5) = 0 or (x − 4) = 0
x = 0, x = 4 or x = 3 
Check the original equation:
x^{ 3} = 9x^{ 2} − 20x 
⇒ x = 0:
⇒ x = 4:
⇒ x = 5 
(0)^{3} = 9(0)^{2} − 20(0) , 0 = 0  0
(4)^{3} = 9(4)^{2} − 20(4) , 64 = 144  80
(5)^{3} = 9(5)^{2} − 20(5) , 125 = 225  100 