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:

Example 1

Factor 9 x ^2 + 45x.

Solution

This example has two common factors 3x and 9x. We will go with the greatest common factor 9 to write the answer in the most simplified form like this:

= 9x (x + 5)

Example 2

Factor 18 x^3 + 12 x^2 + 6x.

Solution

There are three common factors in the above trinomial which are 2x, 3x and 6x. However, we will choose the greatest common factor 6x to simplify the above trinomial.

= 6x (3x ^2 + 2x + 1)

Example 3

Factor 14x ^3 + 42 x ^ 2 + 56x + 70.

Solution

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:

7 (2x ^3 + 6 x ^2 + 8x + 10)

       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:

Example 1

Factor 3x ^3 + 6x ^ 2 + 4xy ^ 2 + 8 y ^ 2

Solution

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 3x ^ 2 and between third and fourth term is 4 y ^ 2. Hence, we will write the factors as:

3x ^2 (x + 2) + 4 y ^2 (x + 2)

You can see that (x + 2) is the common factor in both the groups, hence we will write the final answer in factored form like this:

(3x ^ 2 + 4 y ^2) (x + 2)

Example 2

Factor 5x ^ 3  + 6 x y ^2 + 20 x ^2  + 24 y^ 2.

Solution

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:

5x ^3 + 20 x ^2 + 6 xy ^ 2 + 24 y ^ 2

5x ^2 is the common factor between the first and second terms, and 6 y^ 2 is the common term between the third and fourth terms of the above polynomial.

5x ^ 2 ( x + 4) + 6 y ^2 ( x + 4)

(x + 4) is the common factor in the above polynomial, hence we will write the factored form like this:

( 5 x^2 + 6 y ^ 2) (x + 4)

    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:

m ^ 2 - n ^ 2 = (m - n) (m + n)

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.

Example 1

Factor 25 x ^ 2 - 49 y ^ 2

Solution

The above examples are written as the difference of two squares. It means that the term 25 x^ 2 can be written as 5x \cdot 5x and the second term 49 y ^2 can be written as 7y \cdot 7y. According to the difference of perfect square formula, we will write the above binomial in factored form like this:

( 5x + 7y) (5x - 7y)

The above factors are in their most simplified form.

Example 2

Factor m ^ 4 - n ^ 4.

Solution

The above terms are perfect squares because we can write m ^ 4 in a product form m ^ 2 \cdot m ^2. Similarly, we can write n ^4 as n ^2 \cdot n ^2. Using the difference of perfect squares formula, we will write this binomial as:

(m ^ 2 + n ^2) ( m ^2 - n ^2)

You can see that the second factor (m ^ 2 - n ^2) can be factored again. Hence, we will write the factors as:

(m ^ 2 + n ^2) ( m + n) (m - n)

The above factors are irreducible.

Example 3

Factor 16 x ^ 4 - 81 y ^ 4.

Solution

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:

( 4 x ^ 2 + 9 y^2) ( 4 x ^2 - 9 y ^2)

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:

( 4 x ^ 2 + 9 y^ 2) ( 2x - 3y) (2x + 3y)   <h2>     4. Sum and Differences of Cube</h2> The formula for factoring the sum of cubes is: m ^ 3 + n ^ 3 = ( m + n) (m ^2 - mn + n^2) The formula for factoring the difference of cubes is: m ^3 - n ^3 = (m - n) (m ^2 + n ^ 2 + mn) Consider the following examples to know how to factor the sum and difference of cubes using the above formulas: <h3>Example 1</h3> Factor27 x^3 - 125 y ^3<h3>Solution</h3> Both the terms are perfect cubes. Hence, we will use the following difference of two cubes formula to factor the binomial in the above example. m ^3 - n ^3 = (m - n) (m ^2 + n ^ 2 + mn) Simplifying the above example through substitution of values will yield the following factors: (3x - 5y) (9x ^2 + 15xy +25 y^2) <h3>Example 2</h3> Factor27 x ^3 + 64 y ^ 3<h3>Solution</h3> Both terms in the above example are perfect cubes. We will use the following sum of cubes formula to factor the polynomial in the example: m ^ 3 + n ^ 3 = ( m + n) (m ^2 - mn + n^2) Substitute the values in the above formula to get the factored form: (3x + 4y) ( 9x ^ 2 - 12xy + 16y ^2)   <h2>   5. Factoring Trinomials</h2> When we are given a trinomial, we factor it by expanding the middle term in such a way that the sum or difference of two terms is equal to the midterm and their multiplication is equal to the product of first and third terms. The trinomial ax ^ 2 + bx + cis a quadratic equation. A quadratic equation is a second degree polynomial which means that the highest degree of a polynomial is 2. The final result of this type of factorization is a product of two binomials. However, it is not always necessary that a trinomial is a quadratic function. Consider the following examples to understand this concept further. <h3>Example 1</h3> Factor6x ^ 2 - 11xy - 10y^2. <h3>Solution</h3> The above example is not a quadratic formula, because quadratics have a constant term in the end. Whether or not a trinomial is a quadratic function, we use the same process to factor it. Follow the following procedure to factor the above trinomial: <ul>   <li>Multiply the leading terms and constant together to get -60 x ^2 y ^2.</li>   <li>Expand the middle term in such a way that the sum or difference of the terms is equal to the midterm and the product is equal to -60 x ^2 y ^2. We can write-11xy as-15xy + 4 xy.</li>   <li>Factor by grouping like this.</li> </ul> = 6 x ^ 2 - 15xy + 4xy - 10 y ^2 You can see that-15xy + 4xy = -11xyand-15xy \cdot 4xy = -60 x ^2 y ^2. = 3x ( 2x - 5y ) + 2y (2x - 5y) You can see that the term(2x - 5y)is common, hence we can write the final answer as follows: = (3x + 2y) (2x - 5y) <h3>Example 2</h3> Factor 20 x ^2 + 21x + 4. <h3>Solution</h3> It is an example of a quadratic polynomial because we have constant at the end and the leading coefficient has a degree 2. Follow the following procedure to factor the above trinomial: <ul>   <li>Multiply the first and third terms together to get 80 x ^2 .</li>   <li>Expand the middle term in such a way that the sum or difference of the terms is equal to the midterm and the product is equal to 80 x ^2 . We can write+21x as16 x + 5 x.</li>   <li>Factor the terms by grouping like this.</li> </ul> = 20 x ^2 + 16x + 5x + 4 Factor the terms by grouping together first and the second, and third and the fourth terms together: =  4x ( 5x + 4) + 1 (5x + 4) = (4x + 1) (5x + 4) <h3>Example 3</h3> Factor2x ^2 + 18x + 36. <h3>Solution</h3> Follow the following procedure to factor the above trinomial: <ul>   <li>Multiply the first and third terms together to get 72 x ^2 .</li>   <li>Expand the middle term in such a way that the sum or difference of the terms is equal to the midterm and the product is equal to 72 x ^2 . We can write+18x as12 x + 6x.</li>   <li>Factor the terms by grouping like this.</li> </ul> = 2 x ^2 + 12x + 6x + 36 Now, factor the above polynomial by grouping: = 2x ( x + 6) + 6 ( x + 6) You can see that(x + 6)is the common factor. Hence, we can rewrite the final answer like this: = (2x + 6) (x + 6)$

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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