Let\u2019s look at a few examples.<\/p>\n
Example 1<\/h3>\n
Consider the expression \\(x^2-9\\). Can this expression be factored using the difference of squares? We notice that the first term is raised to the second power, but the second term is not. Can we rewrite 9 so that it IS raised to the second power? We sure can. Let\u2019s rewrite the expression \\(x^2-9\\) as \\(x^2-3^2\\) so that each term is a perfect square. Now the second clue. Is subtraction the symbol between each term? Yes! Both clues are present, so the expression can be factored using the difference of squares. Set up two empty sets of parentheses and break apart the original expression.<\/p>\n
\\(x^2-9\\) becomes \\((x+3)(x-3)\\). Notice that if \\((x+3)(x-3)\\) is multiplied, the product is the original expression \\(x^2-9\\). This is one way to check that you have factored correctly.<\/p>\n
Let\u2019s examine why it is crucial that subtraction is the symbol between both terms. In the previous example, \\(x^2-9\\) was factored into \\((x+3)(x\u22123)\\). When this is multiplied using the FOIL technique (First, Outer, Inner, Last) \\((x+3)(x\u22123)\\) becomes \\(x^2-3x+3x-9\\). As you can see, the two middle terms \\(-3x\\) and \\(3x\\) will cancel out. This leaves only \\(x^2-9\\).<\/p>\n
The original subtraction symbol between the terms is what allows these middle terms to cancel out. If the original expression was \\(x^2+9\\), with addition between the two terms, it would not be possible to factor this using the difference of squares. If we tried to break this apart into \\((x+3)(x+3)\\), the result would be \\(x^2+3x+3x+9\\), which simplifies to \\(x^2+6x+9\\), which does not match the original expression. The important thing to remember is that subtraction between the two terms is required for an expression to be factored using the difference of squares.<\/p>\n
Example 2<\/h3>\n
Let\u2019s look at a more complex example.<\/p>\n
Consider the binomial \\(4x^2-49\\). Can this be factored using the difference of squares? Subtraction is the symbol between both terms, which is a good start. The first term is raised to the power of two which is also good. When we look at the second term, \\(49\\), we notice that this can be written as \\(72\\). Now we have \\(4x^2-7^2\\). All clues indicate that the expression \\(4x^2-49\\) can be rewritten as the difference of two squares. \\(4x^2-49\\) becomes \\((2x+7)(2x-7)\\). We can check this by (FOIL)ing and checking that the product is in fact \\(4x^2-49\\).<\/p>\n
Sometimes a binomial can be factored in one step. For example, \\(x^2-25\\) can be factored simply by splitting the \\(x\\)-squared and the \\(25\\) into \\((x+5)(x\u22125)\\). Many binomials can be factored this way. However, some binomials will require you to pull out a common factor before it is in a form that can be factored using the difference of squares. For example, it appears that the binomial \\(5x^2-45\\) is unable to be factored using the difference of squares because \\(5x^2\\) and \\(45\\) are not perfect squares. However, when the common factor of \\(5\\) is pulled out, the expression becomes \\(5(x^2-9)\\) which has two terms that are perfect squares. From here, \\((x^2-9)\\) can be split up into the factors \\((x+3)(x\u22123)\\), resulting in a final answer of \\(5(x+3)(x\u22123)\\).<\/p>\n
To review, we know that a binomial can be written as the difference of two squares if both terms are perfect squares and they are separated by subtraction. The difference of two squares is a useful theorem because it tells us if a quadratic equation can be written as the product of two binomials.<\/p>\n Here are a few sample questions going over the difference of squares. \\(x^2-16\\)<\/p>\n<\/div>\n\t\t\t\t\t![\"Click](\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/imcr20-New.png\")
<\/a>\nDifference of Squares Sample Questions<\/h2>\n
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\nFactor the binomial below using the difference of two squares method.<\/p>\n