The terms are not perfect squares or perfect cubes. However, notice that they do have acommon factor. First, factor out the GCF,3xy.81x4y+3xy4=3xy(27x3+y3)The resulting binomial factor is a sum of cubes witha=3xandb=y.81x4y+3xy4==3xy(27x3+y3)3xy(3x+y)(9x2−3xy+y2)Answer:3xy(3x+y)(9x2−3xy+y2)When the degree of the special binomial is greater than two, we may need to apply the formulasmultiple times to obtain a complete factorization. A polynomial is completely factoredApolynomial that is prime or written as a product of prime polynomials. when it is prime or iswritten as a product of prime polynomials.Example 12Factor completely:x4−81y4.Solution:First, identify what is being squared.x4−81y4=()2−()2To do this, recall the power rule for exponents,(xm)n=xmn.When exponents are raised to apower, multiply them. With this in mind, we findx4−81y4=(x2)2−(9y2)2Therefore,a=x2 andb=9y2.Substitute into the formula for difference of squares.x4−81y4=(x2+9y2)(x2−9y2)At this point, notice that the factor(x2−9y2)is itself a difference of two squares and thus canbe further factored usinga=x2 andb=3y.The factor(x2+9y2)is prime and cannot befactored using real numbers.x4−81y4==(x2+9y2)(x2−9y2)(x2+9y2)(x+3y)(x−3y)Answer:(x2+9y2)(x+3y)(x−3y)When factoring, always look for resulting factors to factor further.Example 13Factor completely:64x6−y6.Solution:This binomial is both a difference of squares and difference of cubes.64x6−y664x6−y6==(4x2)3−(y2)3Difference of cubes(8x3)2−(y3)2Difference of squares
When confronted with a binomial that is a difference of both squares and cubes, as this is, makeit a rule to factor using difference of squares first. Therefore,a=8x3 andb=y3.Substitute intothe difference of squares formula.64x6−y6=(8x3+y3)(8x3−y3)The resulting two binomial factors are sum and difference of cubes. Each can be factoredfurther. Therefore, we haveThe trinomial factors are prime and the expression is completely factored.Answer:(2x+y)(4x2−2xy+y2)(2x−y)(4x2+2xy+y2)As an exercise, factor the previous example as a difference of cubes first and then compare theresults. Why do you think we make it a rule to factor using difference of squares first?Try this!Factor:a6b6−1Answer:(ab+1)(a2b2−ab+1)(ab−1)(a2b2+ab+1)(click to see video)Key Takeaways●The GCF of two or more monomials is the product of the GCF of the coefficients and thecommon variable factors with the smallest power.●If the terms of a polynomial have a greatest common factor, then factor out that GCFusing the distributive property. Divide each term of the polynomial by the GCF todetermine the terms of the remaining factor.