# How to Factor Polynomials Completely

By Isaiah David

The hardest part about factoring a polynomial completely is knowing when it is finished. Some polynomials can be factored into first order variables and integers. Others can't be reduced nearly so completely. Get a lot of practice with quadratic equations and differences of squares so that by the time you start dealing with third- and fourth-order polynomials, you know how to tell when you have factored an equation as far as it can be factored.

Look for common factors. A common factor is a variable or number that can be factored out of each term in the equation. For example, in the polynomial 2x^3 + 6x + 10, all three terms are even and are therefore divisible by 2. Therefore, 2 is a factor of all 3 terms. In the polynomial 8x^4 + 2x^3 + x^2, x^2 is a factor of all three terms, since each of them contain at least an x^2 term.

Factor out the common factors. In the first example above, you can use the distributive property to factor out the 2:2x^3 + 6x + 10 = 2(x^3 + 3x + 5)In the second example, we can factor out the x^2:8x^4 + 2x^3 + x^2 = x^2(8x^2 + 2x + 1)Sometimes, you can factor out both a number and a variable. For example, in 3x^2 + 6x, you can factor out 3x:3x^2 + 6x = 3x(x + 2)

Look for a sum or difference of cubes. If, after factoring out your all the common factors, you only have a cubed variable and a cubed number left, you either have a difference of cubes or a sum of cubes. If one number is subtracted from another, it is a difference of cubes. If both numbers are added, it is a sum of cubes. For example, the polynomial equation x^4 + 8x can have an x factored out, resulting in x(x^3 + 8). x^3 is a cubed number, and 8 = 2^3. Therefore, you have a sum of cubes.

Plug in the formula for the sum or difference of cubes. The formula for a sum of cubes is:A^3 + B^3 = (A + B)(A^2 - AB + B^2)The formula for a difference of cubes is:A^3 - B^3 = (A - B)(A^2 + AB + B^2)So plugging in the problem from step 3, we get:x^4 + 8x =x(x^3 + 8)x(x^3 + 2^3)x(x + 2)(x^2 - 2x + 4)

Look for a difference of squares and apply the formula. A difference of squares is just like a difference of cubes, except that it involves a factorial with squared terms, such as x^2 - 4 = x^2 - 2^2. The formula is: A^2 - B^2 = (A + B)(A - B). So using that formula, we get:x^2 - 4 =x^2 - 2^2 = (x + 2)(x - 2)

Factor any remaining quadratic equations that can be factored. For example, in the expression x^2 + 7x +10, we need to find two numbers that multiply to 10 and add up to 7. Since 5 * 2 = 10, and 5 + 2 = 7, we get:x^2 + 7x + 10 =(x + 2)(x + 5)