Factoring is a very handy tool to solve a lot of algebraic equations quickly and easily. There are problems in algebra where you require the application of factoring at the very first step of the solution else you won’t be able to proceed. So factoring is a very important chapter of intermediate algebra. Factoring, as you may have a little idea from the lesson regarding quadratic equation, or also from basic mathematics is to break an expression into its factors, that is the product of which would yield the given expression. Also it is important that the polynomial is completely factored, which means the deduced factors can’t be factored anymore.
But reducing long polynomial expressions into its factors is not at all easy and you can’t just always think and calculate in your mind to find the exact factors of a quadratic or higher polynomial equation. That’s why here we will look into few formulae and ways of calculating factors of equations which have their variables raised to powers of square or cube.
Let’s first start with difference of squares.
Difference of squares is an expression or equation that needs us to subtract two values raised to their squares. For example:
X2 – 9 = (X ) (X )
Now, as there is no figure such as ‘ax’ (where a is a constant) in the expression we need to find factors of 9 that sums up to zero, which is basically square roots of a square with opposite signs, here it is +3 and –3. So;
X2 – 9 = (X + 3) (X – 3)
So, to reach a formula here what we need to do is find the square roots of the two values given and multiply their sum and difference, i.e.
a2 – b2 = (a + b) (a – b) or, a – b = (√a + √b) (√a – √b)
Next will be sums and differences of cubes.
Unlike sum of squares, sum of cubes can be factored and also can be based on a simple formula.
X3 + Y3 = (X + Y) (X2 – XY + Y2)
Similarly, difference of cubes can be represented as:
X3 – Y3 = (X – Y) (X2 + XY + Y2)
So, basically if you look at it, the formulae vary only by their signs for sum and difference of cubes. Hence to make it easier we can write;
a3 ± b3 = (a ± b) (a2 Ŧ ab + b2) i.e. (a [same sign] b) (a2 [opposite sign] ab [always positive] b2)
x3 – 27 = x3 – 33 = (x – 3) ( x2 + 3x + 32) = (x – 3) ( x2 + 3x + 9)
8x3 + 64 = (2x)3 + 43 = (2x + 4) ((2x)2 – 2x·4 + 42) = (2x + 4) (4x2 – 8x + 16)
Also, you need to first judge the equation to find out which formula to apply where. There are few possibilities where you can apply both. For example, (x6 – 64) and similar ones which can be solved by both ways.