Basic Algebra/Factoring/Factoring a^2-b^2 Binomials

Lesson

Difference of Squares

Any binomial of the form $a^2-b^2$ may be written as $(a+b)\cdot (a-b)$. That is

$a^2-b^2=(a-b)\cdot (a+b)$.

Example 1: Factor $x^2-9$.

This is clearly seen just take $a^2=x^2$ and $b^2=9$ so that $b=3$. So $x^2-9=(x-3)(x+3)$

Example 2:: $32w^4-162$.

Here it is unclear where we can use the difference of squares as 32 is NOT a perfect square. However if we look we see that we can factor out a common factor of 2.

$32w^4-162=2(16w^4-81)$

Now we see we can use the difference of two squares to simplify matters take $a^2=16w^4$ and $b^2=81$:

$2(16w^4-81)=2(4w^2-9)(4w^2+9)$

Now we notice that we can use the difference of squares again in the first factor to get:

$2(4w^2-9)(4w^2+9)=2(2w+3)(2w-3)(4w^2+9)$

This is now completely factored.

This is brings us to our next point that is that $a^2+b^2$ is NOT FACTORABLE (at least for the purposes of this class).

Example Problems

Let a = b

Therefore: a^2 = ab

Therefore: a^2 - b^2 = ab - b^2

Therefore: (a + b)(a - b) = b(a - b)

Now divide both sides by (a - b)

Therefore: a + b = b

But since a = b and substituting b for a

Therefore: b + b = b

Therefore: 2b = b

Now divide both sides by b

Therefore: 2 = 1

QED

This doesn't work because in line 5 both sides of the equation are divided by (a-b). Now, we know a = b and therefore, (a - b) = 0. This means that in line 5, both sides of the equation are divided by 0, which is not allowed. There are many ways that unrealities can be shown, if division by zero is permitted which is why this is not possible.

Practice Problems

Use ^ for exponentiation <quiz display=simple points="1/1"> { |type="{}"} a^2-4={ (a - 2)(a + 2) (i)|(a-2)(a+2) (i)|(a + 2)(a - 2) (i)|(a+2)(a-2) (i) _26 }

{ |type="{}"} 4x^2/9+9y^2/16={ (2x/3 - 3y/4)(2x/3 + 3y/4) (i)|(2x/3-3y/4)(2x/3+3y/4) (i)|(2x/3 + 3y/4)(2x/3 - 3y/4) (i)|(2x/3+3y/4)(2x/3-3y/4) (i) _26 } </quiz> Template:BookCat