Algebra

Factorisation II

This is when we get closer to quadratic, which in my opinion, is the core of Algebra. Consider the fundamental factorisation rule:

ax + bx = (a + b)x

We can use this rule vice versa as well:

(a + b)x = ax + bx

We can also redefine x to anything. In this example, I'd like to say:

x = a + b

So when we substitute its value we get:

(a + b)x = ax + bx

(a + b)(a + b) = a(a + b) + b(a + b)

Using the fundamental rule we can find a(a + b) and  b(a + b).

a(a + b) = aa + ab = a^2  + ab

b(a + b) = ba + bb  = ab + b^2

So that means:

a(a + b) + b(a + b) = (a^2 + ab) + (ab + b^2)

a(a + b) + b(a + b) = a^2 + 2ab + b^2

We can substitute this to get:

(a + b)(a + b) = a(a + b) + b(a + b)

(a + b)(a + b) = a^2 + 2ab + b^2

We could also use a similar method to figure out:

(a + b)(a - b)

a(a - b) + b(a - b)

a^2 - ab + ab - b^2

a^2 - b^2

And so our final result is:

(a + b)(a - b) = a^2 - b^2

So remember these two rules, as they pop up quite regularly:

(a + b)(a - b) = a^2 - b^2

(a + b)(a + b) = a^2 + 2ab + b^2

Now let's use these rules in actual examples:

Challenge 1: (x + 2)^2 = x^2 + (x + 1)^2

Answer:

Take a deep breath, because this is going to be big one.

(x + 2)^2 = x^2 + (x + 1)^2

(x + 2)^2 - (x + 1)^2 = x^2 + (x + 1)^2 - (x + 1)^2

(x + 2)^2 - (x + 1)^2 = x^2

Using our rule:

(a + b)(a - b) = a^2 - b^2

Substitute a = x + 2, and b = x + 1 to get:

(x + 2 + x + 1)(x + 2 - (x + 1)) = (x + 2)^2 - (x + 1)^2

(2x + 3)(1) = (x + 2)^2 - (x + 1)^2

2x + 3 = (x + 2)^2 - (x + 1)^2

(x + 2)^2 - (x + 1)^2 = 2x + 3

From the previous equation we can say:

(x + 2)^2 - (x + 1)^2 = x^2

2x + 3 = x^2

2x + 3 - x^2 = x^2 - x^2

- x^2 + 2x + 3 = 0

-1(- x^2 + 2x + 3) = -1(0)

x^2 - 2x - 3 = 0

x^2 - 2x - 3 - 4x + 12 = 0 - 4x + 12

x^2 - 2x - 4x + 12 -3 = - 4x + 12

x^2 - 6x + 9 = - 4x + 12

From the (a + b)(a + b) rule we know:

a^2 + 2ab + b^2 = (a + b)(a + b)

We can substitute a = x and b = -3

a^2 + 2ab + b^2 = (a + b)(a + b)

x^2 + (- 6x) + 3^2 = (x + (- 3))(x + (-3))

x^2 - 6x + 9 = (x - 3)(x - 3)

From the previous equation:

x^2 - 6x + 9 = - 4x + 12

(x - 3)(x - 3) = - 4x + 12

(x - 3)(x - 3) = - 4(x - 3)

(x - 3)(x - 3) + 4(x - 3) = - 4(x - 3) + 4(x - 3)

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

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

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

(x + 1)(x - 3) = 0

So it is either x + 1 = 0 or x - 3 = 0 in order to satisfy the equation.

(1st Option) x + 1 = 0

                           x = - 1  

(2nd Option) x - 3 = 0

                            x  = 3

So there are two solutions: x = -1 and x = 3.

It is okay if you didn't understand the challenge. All you need to understand is these rules:

(a + b)(a - b) = a^2 - b^2

(a + b)(a + b) = a^2 + 2ab + b^2

I might add more parts when I feel like. Thank you for reading.