{"id":70271,"date":"2021-03-11T14:08:14","date_gmt":"2021-03-11T20:08:14","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=70271"},"modified":"2026-03-28T12:21:28","modified_gmt":"2026-03-28T17:21:28","slug":"factoring-polynomials","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/factoring-polynomials\/","title":{"rendered":"Factoring Polynomials Overview"},"content":{"rendered":"

Factoring polynomials is an important skill to master because it allows us to rewrite polynomials in a simpler form. The process of factoring helps us understand more about the equations we are working with and produces useful information. For example, solutions are often much easier to identify when a polynomial is in a factored form. Factoring polynomials also makes it easier to graph an equation.<\/p>\n

Factoring Polynomials Sample Questions<\/a><\/div>\n

There are many different factoring techniques. The most common strategy for factoring polynomials is to simply factor out the greatest common factor. If there is no clear factor in common, then another approach needs to be implemented. Another common approach is to split the polynomial into two sets of parentheses that are multiplied by each other.<\/p>\n

Strategy 1<\/h3>\n

Let\u2019s take a closer look at the first strategy: factoring out the greatest common factor.<\/p>\n

Consider the polynomial \\(6x^2+9x\\). We can simplify this polynomial by pulling out the greatest common factor. The largest factor that is shared by both terms is \\(3x\\). Factor out the \\(3x\\) by dividing each term by \\(3x\\). \\(6x^2\\) becomes \\(3x\\) and \\(9x\\) becomes 3. So 6×2+9x becomes 3x(x+3). Notice that this factored form will multiply back to its original form. In this case, \\(3x\\) times \\((x+3)\\) gives us \\(6x^2+9x\\),\u00a0 which proves that the factored form is equivalent to the original polynomial.<\/p>\n

Strategy 2<\/h3>\n

Let\u2019s take a look at a tougher polynomial that has three terms.<\/p>\n

Consider the following polynomial: \\(4x^3-2x^2+6x\\). Look at each term and determine if there is a common factor shared by all terms. In this example, the greatest common factor<\/a> is \\(2x\\). Now \u201cfactor this out\u201d by dividing each term by \\(2x\\). \\(4x^3-2x^2+6x\\) becomes \\(2x(2x^2-x+3)\\). We can check our work to make sure that we have factored correctly by multiplying \\(2x\\) by \\((2x-x+3)\\). If the product is the original polynomial, we have factored correctly. \\(2x\\) multiplied by \\((2x^2-x+3)\\) equals \\(4x^3-2x^2+6x\\), so we have factored correctly.<\/p>\n

It is convenient when polynomials have a clear common factor. However, many times there will not be an obvious factor that is shared by all terms. For example, \\(x^2+5x+6\\). There are no factors that are shared by all terms, so we need to approach this polynomial in a different way. This example can be factored using the approach of splitting the polynomial into two sets of parentheses that are multiplied by each other. Let\u2019s use this approach with the polynomial \\(x^2+5x+6\\).<\/p>\n

It’s true to say that \\(x^2+5x+6\\) has no common factors, so set up two empty sets of parentheses that are multiplied by each other \\(( )( )\\). Now, break apart the first term. \\(x^2\\) can be split up into x times x, so place this in the parentheses \\((x\\text{ }\\text{ }) (x\\text{ }\\text{ })\\). Now take a close look at the next two terms \\(5x+6\\). Ask yourself, \u201cWhat multiplies<\/em> to 6 and adds<\/em> to 5?\u201d In this case, 2 times 3 multiplies to 6, and 2 plus 3 adds to 5. Place 2 and 3 in the parentheses. \\((x\\text{ }\\text{ }) (x\\text{ }\\text{ })\\)\u00a0becomes \\((x+2)(x+3)\\). This is the factored version of the original polynomial. Again, we can check that we have factored correctly by multiplying \\((x+2)\\) times \\((x+3)\\) to make sure that our product results in the original polynomial.<\/p>\n

Strategy 3<\/h3>\n

Let\u2019s take a look at a polynomial that can be factored by pulling out the greatest common factor AND by splitting up a polynomial into two sets of parentheses that are multiplied.<\/p>\n

For example, let\u2019s look at the polynomial \\(12x^2-27\\). When we look for a common factor, we notice that both terms are divisible by 3. 3 is the largest factor shared by both terms, so let\u2019s factor this out. \\(12x^2-27\\) becomes \\(3(4x^2-9)\\). Many polynomials are successfully factored after just this single step. However, sometimes it takes multiple steps to completely<\/em> factor a polynomial. \\(3(4×62-9)\\) is close to complete, but consider what is inside the parentheses. \\((4x^2-9)\\) can be broken down even further. \\((4x^2-9)\\)\u00a0can be split up into two sets of parentheses that are multiplied by each other. \\((x^2-9)\\) becomes \\((2x\u22123)(2x+3)\\). Now we are left with \\(3x(2x\u22123)(2x+3)\\), which is factored completely.<\/p>\n

Remember, when factoring a polynomial, you are simply rewriting it in a different way. The value has not changed at all, just the form. The objective of factoring is to break down a long and complicated polynomial into smaller pieces that are more manageable. This makes it easier to identify solutions, which inevitably makes the polynomial easier to graph.<\/div>\n\"Click<\/a>\n

Factoring Polynomials Problems<\/h2>\n

Here are a few sample questions going over factoring polynomials.
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\n\t\t\t\t\tQuestion #1:<\/strong>\n\t\t\t\t\t

Factor the following polynomial: \\(2x-4y+2\\)<\/p>\n<\/div>\n\t\t\t\t\t

\\(4(x-2y+2)\\)<\/div>
\\(2(x-2y)\\)<\/div>
\\(4(x-2y)\\)<\/div>
\\(2(x-2y+1)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t