If you have a binomial where both terms are a perfect square, you can factor it using difference of squares.
This also works when you have a number plus another number.
I'll provide two examples:
(x^2 - 9)
The difference of squares states: (a^2 - b^2) = (a + b)(a - b)
In this case, (x^2 - 9) = (x + 3)(x - 3)
We can also apply the difference of squares with a number plus another.
(x^2 + 25)
We can rewrite this binomial as: (x^2 - (-25))
Now, we can apply the same steps to factor.
(x^2 - (-25)) = (x + (√-25))(x - (√-25))
Because we have √-25, we can simplify it by multiplying it by i, which will remove the negative.
This leaves us with (x + 5i)(x - 5i), which is the factored form of x^2 + 25.
We can verify this by using FOIL.
x^2 - 5ix + 5ix - 25i^2
x^2 + 25i^2
i^2 can be interpreted as -1, and so we can change the + to -
x^2 - 25
