January 25, 2020
Factoring and multiplying polynomials - both processes are inverses of each other. If we divide a polynomial by any of its factors, we will get the remainder zero. Before factoring a polynomial equation, it is better to rewrite it in descending order. We follow the following procedure while factoring a polynomial:
- First, we need to find the greatest common factor in the polynomial.
- We see if the polynomial can be factored through grouping
- We look if the polynomial can be factored by using the difference of squares and sum or difference of cube formula.
- If the polynomial is a trinomial and cannot be factored using any of the above methods, then we factor it by following a special procedure which involves expanding the middle term and then finding the factors through grouping.
We find zeroes or roots of a polynomial after factorization. Sometimes, we have to factor a polynomial twice because to write an equation in its most simplified form, we have to list down all the possible factors.
Methods for Factoring a Polynomial
Generally, we use the following five methods for factoring the polynomial functions.
- Greatest common factor (G.C.F)
- Factoring by grouping
- Differences of perfect squares
- Sum and difference of cubes
- Factoring trinomials
Lets us discuss the above methods of factoring a polynomial one by one.
1. Greatest Common Factor (G.C.F)
We always look for the greatest common factor before factoring the trinomials. Consider the examples below:
This example has two common factors and . We will go with the greatest common factor 9 to write the answer in the most simplified form like this:
There are three common factors in the above trinomial which are , and . However, we will choose the greatest common factor to simplify the above trinomial.
The above polynomial function has two common factors, 2 and 7. We will use the greatest common factor 7 to factor the polynomial like this:
2. Factoring by grouping
Another method of factoring a polynomial is grouping. Sometimes, there is no common factor in all the terms of a polynomial. A separate group of terms has a common factor among them, so we group an even number of terms together and find the common factor between them. Consider the following examples:
You can see that in the above polynomial, there is no common factor. Instead, we can group first and the second, and third and fourth like terms together and find the common factor between them.
The common factor between the first and second term is and between third and fourth term is . Hence, we will write the factors as:
You can see that is the common factor in both the groups, hence we will write the final answer in factored form like this:
You can see that there are no common factors among all terms in the above polynomial function. Hence, we will see that either we can simplify the above polynomial by grouping the terms together or not. If you look at the above polynomial carefully, you will get to know that the first and third terms and the second and fourth terms have the common factors between them. Hence, we will write the polynomial like this:
is the common factor between the first and second terms, and is the common term between the third and fourth terms of the above polynomial.
is the common factor in the above polynomial, hence we will write the factored form like this:
3. The Difference of Perfect Squares
If there are two terms which are perfect squares and there is a negative sign between the terms, then we can write them in factored form as:
When we take the square root of a perfect square, we get a whole number instead of a decimal point. Examples of perfect squares include 25, 36, 81 and 100 etc.
Consider the following examples which will clarify the above concept.
The above examples are written as the difference of two squares. It means that the term can be written as and the second term can be written as . According to the difference of perfect square formula, we will write the above binomial in factored form like this:
The above factors are in their most simplified form.
The above terms are perfect squares because we can write in a product form . Similarly, we can write as . Using the difference of perfect squares formula, we will write this binomial as:
You can see that the second factor can be factored again. Hence, we will write the factors as:
The above factors are irreducible.
Both the terms in the above question are perfect squares and there is also a negative sign between them. Hence, we can write the above binomial by using the difference of perfect squares formula like this:
Notice that the second factor can be simplified further because both the terms inside the parentheses are perfect squares and we can write them as a product of factors like this:
m ^ 3 + n ^ 3 = ( m + n) (m ^2 - mn + n^2)m ^3 - n ^3 = (m - n) (m ^2 + n ^ 2 + mn)27 x^3 - 125 y ^3m ^3 - n ^3 = (m - n) (m ^2 + n ^ 2 + mn) (3x - 5y) (9x ^2 + 15xy +25 y^2)27 x ^3 + 64 y ^ 3 m ^ 3 + n ^ 3 = ( m + n) (m ^2 - mn + n^2) (3x + 4y) ( 9x ^ 2 - 12xy + 16y ^2) ax ^ 2 + bx + c6x ^ 2 - 11xy - 10y^2 -60 x ^2 y ^2 -60 x ^2 y ^2-11xy -15xy + 4 xy = 6 x ^ 2 - 15xy + 4xy - 10 y ^2-15xy + 4xy = -11xy-15xy \cdot 4xy = -60 x ^2 y ^2 = 3x ( 2x - 5y ) + 2y (2x - 5y)(2x - 5y) = (3x + 2y) (2x - 5y) 20 x ^2 + 21x + 4 80 x ^2 80 x ^2 +21x 16 x + 5 x = 20 x ^2 + 16x + 5x + 4= 4x ( 5x + 4) + 1 (5x + 4) = (4x + 1) (5x + 4)2x ^2 + 18x + 36 72 x ^2 72 x ^2 +18x 12 x + 6x = 2 x ^2 + 12x + 6x + 36 = 2x ( x + 6) + 6 ( x + 6)(x + 6) = (2x + 6) (x + 6)$