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## 4.2 Factoring Polynomials

### Learning Objectives

1. Determine the greatest common factor (GCF) of monomials.
2. Factor out the GCF of a polynomial.
3. Factor a four-term polynomial by grouping.
4. Factor special binomials.

## Determining the GCF of Monomials

The process of writing a number or expression as a product is called factoringThe process of writing a number or expression as a product.. If we write the monomial $8x7=2x5⋅4x2$, we say that the product $2x5⋅4x2$ is a factorizationAny combination of factors, multiplied together, resulting in the product. of $8x7$ and that $2x5$ and $4x2$ are factorsAny of the numbers or expressions that form a product.. Typically, there are many ways to factor a monomial. Some factorizations of $8x7$ follow:

$8x7=2x5⋅4x28x7=8x6⋅x8x7=2x⋅2x2⋅2x4} Factorizations of 8x7$

Given two or more monomials, it will be useful to find the greatest common monomial factor (GCF)The product of the common variable factors and the GCF of the coefficients. of each. The GCF of the monomials is the product of the common variable factors and the GCF of the coefficients.

### Example 1

Find the GCF of $25x7y2z$ and $15x3y4z2.$

Solution:

Begin by finding the GCF of the coefficients. In this case, $25=5⋅5$ and $15=3⋅5.$ It should be clear that

$GCF(25,15)=5$

Next determine the common variable factors with the smallest exponents.

$25x7y2z and 15x3y4z2$

The common variable factors are $x3$, $y2$, and $z.$ Therefore, given the two monomials,

$GCF=5x3y2z$

Answer: $5x3y2z$

It is worth pointing out that the GCF divides both expressions evenly.

$25x7y2z 5x3y2z=5x4and15x3y4z25x3y2z=3y2z$

Furthermore, we can write the following:

$25x7y2z =5x3y2z⋅5x4and15x3y4z2=5x3y2z⋅3y2z$

The factors $5x4$ and $3y2z$ share no common monomial factors other than 1; they are relatively primeExpressions that share no common factors other than 1..

### Example 2

Determine the GCF of the following three expressions: $12a5b2(a+b)5$, $60a4b3c (a+b)3$, and $24a2b7c3(a+b)2.$

Solution:

Begin by finding the GCF of the coefficients. To do this, determine the prime factorization of each and then multiply the common factors with the smallest exponents.

$12=22⋅360=22⋅3⋅524=23⋅3$

Therefore, the GCF of the coefficients of the three monomials is

$GCF(12,60,24)=22⋅3=12$

Next, determine the common factors of the variables.

$12a5b2(a+b)5and60a4b3c (a+b)3and24a2b7c3(a+b)2$

The variable factors in common are $a2$, $b2$, and $(a+b)2.$ Therefore,

$GCF=12⋅a2⋅b2⋅(a+b)2$

Note that the variable c is not common to all three expressions and thus is not included in the GCF.

Answer: $12a2b2(a+b)2$

## Factoring out the GCF

The application of the distributive property is the key to multiplying polynomials. For example,

$6xy2(2xy+1)=6xy2⋅2xy+6xy2⋅1Multiplying=12x2y3 + 6xy2$

The process of factoring a polynomial involves applying the distributive property in reverse to write each polynomial as a product of polynomial factors.

$a(b+c)=ab+acMultiplyingab+ac=a(b+c)Factoring$

Consider factoring the result of the opening example:

$12x2y3 + 6xy2=6xy2⋅2xy + 6xy2⋅1Factoring=6xy2( ? )=6xy2(2xy+1)$

We see that the distributive property allows us to write the polynomial $12x2y3 + 6xy2$ as a product of the two factors $6xy2$ and $(2xy+1).$ Note that in this case, $6x2y$ is the GCF of the terms of the polynomial.

$GCF(12x2y3, 6xy2)=6xy2$

Factoring out the greatest common factor (GCF)The process of rewriting a polynomial as a product using the GCF of all of its terms. of a polynomial involves rewriting it as a product where a factor is the GCF of all of its terms.

$8x3+4x2−16x=4x(2x2+x−4)9ab2−18a2b−3ab=3ab(3b−6a−1)}Factoring out the GCF$

To factor out the GCF of a polynomial, we first determine the GCF of all of its terms. Then we can divide each term of the polynomial by this factor as a means to determine the remaining factor after applying the distributive property in reverse.

### Example 3

Factor out the GCF: $18x7−30x5+6x3.$

Solution:

In this case, the GCF(18, 30, 6) = 6, and the common variable factor with the smallest exponent is $x3.$ The GCF of the polynomial is $6x3.$

$18x7−30x5+6x3=6x3(?)$

The missing factor can be found by dividing each term of the original expression by the GCF.

$18x76x3=3x4 −30x56x3=−5x2 +6x36x3=+1$

Apply the distributive property (in reverse) using the terms found in the previous step.

$18x7−30x5+6x3=6x3(3x4−5x2+1)$

If the GCF is the same as one of the terms, then, after the GCF is factored out, a constant term 1 will remain. The importance of remembering the constant term becomes clear when performing the check using the distributive property.

$6x3(3x4−5x2+1)=6x3⋅3x4−6x3⋅5x2+6x3⋅1=18x7−30x5+6x3✓$

Answer: $6x3(3x4−5x2+1)$

### Example 4

Factor out the GCF: $27x5y5z+54x5yz−63x3y4.$

Solution:

The GCF of the terms is $9x3y.$ The last term does not have a variable factor of z, and thus z cannot be a part of the greatest common factor. If we divide each term by $9x3y$, we obtain

$27x5y5z9x3y=3x2y4z 54x5yz9x3y=6x2z −63x3y49x3y=−7y3$

and can write

$27x5y5z+54x5yz−63x3y4=9x3y(?)=9x3y (3x2y4z+6x2z−7y3)$

Answer: $9x3y (3x2y4z+6x2z−7y3)$

Try this! Factor out the GCF: $12x3y4−6x2y3−3xy2$

Answer: $3xy2(4x2y2−2xy−1)$

Of course, not every polynomial with integer coefficients can be factored as a product of polynomials with integer coefficients other than 1 and itself. If this is the case, then we say that it is a prime polynomialA polynomial with integer coefficients that cannot be factored as a product of polynomials with integer coefficients other than 1 and itself.. For example, a linear factor such as $10x−9$ is prime. However, it can be factored as follows:

$10x−9=x(10−9x)or10x−9=5(2x−95)$

If an x is factored out, the resulting factor is not a polynomial. If any constant is factored out, the resulting polynomial factor will not have integer coefficients. Furthermore, some linear factors are not prime. For example,

$5x−10=5(x−2)$

In general, any linear factor of the form $ax+b$, where a and b are relatively prime integers, is prime.

## Factoring by Grouping

In this section, we outline a technique for factoring polynomials with four terms. First, review a preliminary example where the terms have a common binomial factor.

### Example 5

Factor: $7x(3x−2)−(3x−2).$

Solution:

Begin by rewriting the second term $−(3x−2)$ as $−1(3x−2).$ Next, consider $(3x−2)$ as a common binomial factor and factor it out as follows:

$7x(3x−2)−(3x−2)=7x(3x−2)−1(3x−2)=(3x−2)( ? )=(3x−2)(7x−1)$

Answer: $(3x−2)(7x−1)$

Factoring by groupingA technique for factoring polynomials with four terms. is a technique that enables us to factor polynomials with four terms into a product of binomials. This involves an intermediate step where a common binomial factor will be factored out. For example, we wish to factor

$3x3−12x2+2x−8$

Begin by grouping the first two terms and the last two terms. Then factor out the GCF of each grouping: In this form, the polynomial is a binomial with a common binomial factor, $(x−4).$

$=(x−4)( ? )=(x−4)(3x2+2)$

Therefore,

$3x3−12x2+2x−8=(x−4)(3x2+2)$

We can check by multiplying.

$(x−4)(3x2+2)=3x3+2x−12x2−8=3x3−12x2+2x−8 ✓$

### Example 6

Factor by grouping: $24a4−18a3−20a+15.$

Solution:

The GCF for the first group is $6a3.$ We have to choose 5 or −5 to factor out of the second group. Factoring out +5 does not result in a common binomial factor. If we choose to factor out −5, then we obtain a common binomial factor and can proceed. Note that when factoring out a negative number, we change the signs of the factored terms. Answer: $(4a−3)(6a3−5).$ Check by multiplying; this is left to the reader as an exercise.

Sometimes we must first rearrange the terms in order to obtain a common factor.

### Example 7

Factor: $ab−2a2b+a3−2b2.$

Solution:

Simply factoring the GCF out of the first group and last group does not yield a common binomial factor. We must rearrange the terms, searching for a grouping that produces a common factor. In this example, we have a workable grouping if we switch the terms $a3$ and $ab.$ Answer: $(a−2b)(a2+b)$

Try this! Factor: $x3−x2y−xy+y2.$

Answer: $(x−y)(x2−y)$

Not all factorable four-term polynomials can be factored with this technique. For example,

$3x3+5x2−x+2$

This four-term polynomial cannot be grouped in any way to produce a common binomial factor. Despite this, the polynomial is not prime and can be written as a product of polynomials. It can be factored as follows:

$3x3+5x2−x+2=(x+2)(3x2−x+1)$

Factoring such polynomials is something that we will learn to do as we move further along in our study of algebra. For now, we will limit our attempt to factor four-term polynomials to using the factor by grouping technique.

## Factoring Special Binomials

A binomial is a polynomial with two terms. We begin with the special binomial called difference of squares$a2−b2=(a+b)(a−b),$ where a and b represent algebraic expressions.:

$a2−b2=(a+b)(a−b)$

To verify the above formula, multiply.

$(a+b)(a−b)=a2−ab+ba−b2=a2−ab+ab−b2=a2−b2$

We use this formula to factor certain special binomials.

### Example 8

Factor: $x2−9y2.$

Solution:

Identify the binomial as difference of squares and determine the square factors of each term. Here we can write

$x2−9y2=( x )2−( 3y )2$

Substitute into the difference of squares formula where $a=x$ and $b=3y.$

$a2−b2=(a+b)(a−b)↓↓↓↓x2−9y2=(x+3y)(x−3y)$

Multiply to check.

$(x+3y)(x−3y)=x2−3xy+3yx−9y2=x2−3xy+3xy−9y2=x2−9y2 ✓$

Answer: $(x+3y)(x−3y)$

### Example 9

Factor: $x2−(2x−1)2.$

Solution:

First, identify this expression as a difference of squares.

$x2−(2x−1)2=( x )2−( 2x−1 )2$

Use $a=x$ and $b=2x−1$ in the formula for a difference of squares and then simplify.

$a2−b2=(a+b)(a−b)$

$x2−(2x−1)2=[ x+(2x−1)][x−(2x−1)]=( x+2x−1)(x−2x+1)=(3x−1)(−x+1)$

Answer: $(3x−1)(−x+1)$

Given any real number b, a polynomial of the form $x2+b2$ is prime. Furthermore, the sum of squares$a2+b2,$ where a and b represent algebraic expressions. This does not have a general factored equivalent. $a2+b2$ does not have a general factored equivalent. Care should be taken not to confuse this with a perfect square trinomial.

$(a+b)2=(a+b)(a+b)=a2+ab+ba+b2=a2+2ab+b2$

Therefore,

$(a+b)2≠a2+b2$

For example, the sum of squares binomial $x2+9$ is prime. Two other special binomials of interest are the sum$a3+b3=(a+b)(a2−ab+b2)$, where a and b represent algebraic expressions. and difference of cubes$a3−b3=(a−b)(a2+ab+b2)$, where a and b represent algebraic expressions.:

$a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)$

We can verify these formulas by multiplying.

$(a+b)(a2−ab+b2)=a3−a2b+ab2+a2b−ab2+b3=a3+b3 ✓$

$(a−b)(a2+ab+b2)=a3+a2b+ab2−a2b−ab2−b3=a3−b3 ✓$

The process for factoring sums and differences of cubes is very similar to that of differences of squares. We first identify a and b and then substitute into the appropriate formula. The separate formulas for the sum and difference of cubes allow us to always choose a and b to be positive.

### Example 10

Factor: $x3−8y3.$

Solution:

First, identify this binomial as a difference of cubes. Next, identify what is being cubed.

$x3−8y3=(x)3−(2y)3$

In this case, $a=x$ and $b=2y.$ Substitute into the difference of cubes formula.

$a3+b3=(a−b)(a2 + a ⋅ b+b2)↓↓↓↓ ↓↓x3−8y3=(x−2y)((x)2+x⋅2y+(2y)2)=(x−2y)(x2+2xy+4y2)$

We can check this factorization by multiplying.

$(x−2y)(x2+2xy+4y2)=x3+2x2y+4xy2−2x2y−4xy2−8y3=x3 + 2x2y + 4xy2−2x2y−4xy2−8y3=x3−8y3 ✓$

Answer: $(x−2y)(x2+2xy+4y2)$

It may be the case that the terms of the binomial have a common factor. If so, it will be difficult to identify it as a special binomial until we first factor out the GCF.

### Example 11

Factor: $81x4y+3xy4.$

Solution:

The terms are not perfect squares or perfect cubes. However, notice that they do have a common factor. First, factor out the GCF, $3xy.$

$81x4y+3xy4=3xy(27x3+y3)$

The resulting binomial factor is a sum of cubes with $a=3x$ and $b=y.$

$81x4y+3xy4=3xy(27x3+y3)=3xy(3x+y)(9x2−3xy+y2)$

Answer: $3xy(3x+y)(9x2−3xy+y2)$

When the degree of the special binomial is greater than two, we may need to apply the formulas multiple times to obtain a complete factorization. A polynomial is completely factoredA polynomial that is prime or written as a product of prime polynomials. when it is prime or is written as a product of prime polynomials.

### Example 12

Factor completely: $x4−81y4.$

Solution:

First, identify what is being squared.

$x4−81y4=( )2−( )2$

To do this, recall the power rule for exponents, $(xm)n=xmn.$ When exponents are raised to a power, multiply them. With this in mind, we find

$x4−81y4=( x2 )2−( 9y2 )2$

Therefore, $a=x2$ and $b=9y2.$ 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 can be further factored using $a=x2$ and $b=3y.$ The factor $(x2+9y2)$ is prime and cannot be factored 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 13

Factor completely: $64x6−y6.$

Solution:

This binomial is both a difference of squares and difference of cubes.

$64x6−y6=( 4x2 )3−( y2 )3 Difference of cubes64x6−y6=( 8x3 )2−( y3 )2 Difference of squares$

When confronted with a binomial that is a difference of both squares and cubes, as this is, make it a rule to factor using difference of squares first. Therefore, $a=8x3$ and $b=y3.$ Substitute into the difference of squares formula.

$64x6−y6=(8x3+y3)(8x3−y3)$

The resulting two binomial factors are sum and difference of cubes. Each can be factored further. Therefore, we have The 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 the results. Why do you think we make it a rule to factor using difference of squares first?

Try this! Factor: $a6b6−1$

Answer: $(ab+1)(a2b2−ab+1)(ab−1)(a2b2+ab+1)$

### Key Takeaways

• The GCF of two or more monomials is the product of the GCF of the coefficients and the common variable factors with the smallest power.
• If the terms of a polynomial have a greatest common factor, then factor out that GCF using the distributive property. Divide each term of the polynomial by the GCF to determine the terms of the remaining factor.
• Some four-term polynomials can be factored by grouping the first two terms and the last two terms. Factor out the GCF of each group and then factor out the common binomial factor.
• When factoring by grouping, you sometimes have to rearrange the terms to find a common binomial factor. After factoring out the GCF, the remaining binomial factors must be the same for the technique to work.
• When factoring special binomials, the first step is to identify it as a sum or difference. Once we identify the binomial, we then determine the values of a and b and then substitute into the appropriate formula.
• If a binomial is both a difference of squares and cubes, then first factor it as a difference of squares.

### Part A: Factoring out the GCF

Determine the GCF of the given expressions.

1. $9x5$, $27x2$, $15x7$

2. $20y4$, $12y7$, $16y3$

3. $50x2y3$, $35xy3$, $10x3y2$

4. $12x7y2$, $36x4y2$, $18x3y$

5. $15a7b2c5$, $75a7b3c$, $45ab4c3$

6. $12a6b3c2$, $48abc3$, $125a2b3c$

7. $60x2(2x−1)3$, $42x(2x−1)3$, $6x3(2x−1)$

8. $14y5(y−8)2$, $28y2(y−8)$, $35y(y−8)3$

9. $10a2b3(a+b)5$, $48a5b2(a+b)2$, $26ab5(a+b)3$

10. $45ab7(a−b)7$, $36a2b2(a−b)3$, $63a4b3(a−b)2$

Determine the missing factor.

1. $18x4−6x3+2x2=2x2( ? )$

2. $6x5−9x3−3x=3x(​​​​​ ? )$

3. $−10y6+6y4−4y2=−2y2( ? )$

4. $−27y9−9y6+3y3=−3y3( ? )$

5. $12x3y2−8x2y3+8xy=4xy( ? )$

6. $10x4y3−50x3y2+15x2y2=5xy( ? )$

7. $14a4b5−21a3b4−7a2b3=7a2b3( ? )$

8. $15a5b4+9a4b2−3a2b=3a2b( ? )$

9. $x3n+x2n+xn=xn( ? )$

10. $y4n+y3n−y2n=y2n( ? )$

Factor out the GCF.

1. $12x4−16x3+4x2$

2. $15x5−10x4−5x3$

3. $20y8+28y6+40y3$

4. $18y7−24y5−30y3$

5. $2a4b3−6a3b2+8a2b$

6. $28a3b3−21a2b4−14ab5$

7. $2x3y5−4x4y4+x2y3$

8. $3x5y−2x4y2+x3y3$

9. $5x2(2x+3)−3(2x+3)$

10. $y2(y−1)+9(y−1)$

11. $9x2(3x−1)+(3x−1)$

12. $7y2(5y+2)−(5y+2)$

13. $x5n−x3n+xn$

14. $y6n−y3n−y2n$

### Part B: Factoring by Grouping

Factor by grouping.

1. $2x3+3x2+2x+3$

2. $5x3+25x2+x+5$

3. $6x3−3x2+4x−2$

4. $3x3−2x2−15x+10$

5. $x3−x2−3x+3$

6. $6x3−15x2−2x+5$

7. $2x3+7x2−10x−35$

8. $3x3−x2+24x−8$

9. $14y4+10y3−7y−5$

10. $5y4+2y3+20y+8$

11. $x4n+x3n+2xn+2$

12. $x5n+x3n+3x2n+3$

13. $x3−x2y+xy2−y3$

14. $x3+x2y−2xy2−2y3$

15. $3x3y2+9x2y3−x−3y$

16. $2x3y3−x2y3+2x−y$

17. $a2b−4ab2−3a+12b$

18. $a2b+3ab2+5a+15b$

19. $a4+a2b3+a2b+b4$

20. $a3b+2a2+3ab4+6b3$

21. $3ax+10by−5ay−6bx$

22. $a2x−5b2y−5a2y+b2x$

23. $x4y2−x3y3+x2y4−xy5$

24. $2x5y2+4x4y2+18x3y+36x2y$

25. $a5b2+a4b4+a3b3+a2b5$

26. $3a6b+3a5b2+9a4b2+9a3b3$

### Part C: Factoring Special Binomials

Factor.

1. $x2−64$

2. $x2−100$

3. $9−4y2$

4. $25−y2$

5. $x2−81y2$

6. $x2−49y2$

7. $a2b2−4$

8. $1−9a2b2$

9. $a2b2−c2$

10. $4a2−b2c2$

11. $x4−64$

12. $36−y4$

13. $(2x+5)2−x2$

14. $(3x−5)2−x2$

15. $y2−(y−3)2$

16. $y2−(2y+1)2$

17. $(2x+5)2−(x−3)2$

18. $(3x−1)2−(2x−3)2$

19. $x4−16$

20. $81x4−1$

21. $x4y4−1$

22. $x4−y4$

23. $x8−y8$

24. $y8−1$

25. $x2n−y2n$

26. $x2ny2n−4$

27. $x4n−y4n$

28. $x4ny4n−16$

29. $x3−27$

30. $8x3−125$

31. $8y3+27$

32. $64x3+343$

33. $x3−y3$

34. $x3+y3$

35. $8a3b3+1$

36. $27a3−8b3$

37. $x3y3−125$

38. $216x3+y3$

39. $x3+(x+3)3$

40. $y3−(2y−1)3$

41. $(2x+1)3−x3$

42. $(3y−5)3−y3$

43. $x3n−y3n$

44. $x3n+y3n$

45. $a6+64$

46. $64a6−1$

47. $x6−y6$

48. $x6+y6$

49. $x6n−y6n$

50. $x6n+y6n$

51. Given $f(x)=2x−1$, show that $(f+f)(x)=2f(x).$

52. Given $f(x)=x2−3x+2$, show that $(f+f)(x)=2f(x).$

53. Given $f(x)=mx+b$, show that $(f+f)(x)=2f(x).$

54. Given $f(x)=ax2+bx+c$, show that $(f+f)(x)=2f(x).$

55. Given $f(x)=ax2+bx+c$, show that $(f−f)(x)=0.$

56. Given $f(x)=mx+b$, show that $(f−f)(x)=0.$

### Part D: Discussion Board

1. What can be said about the degree of a factor of a polynomial? Give an example.

2. If a binomial falls into both categories, difference of squares and difference of cubes, which would be best to use for factoring, and why? Create an example that illustrates this situation and factor it using both formulas.

3. Write your own examples for each of the three special types of binomial. Factor them and share your results.

1. $3x2$

2. $5xy2$

3. $15ab2c$

4. $6x(2x−1)$

5. $2ab2(a+b)2$

6. $(9x2−3x+1)$

7. $(5y4−3y2+2)$

8. $(3x2y−2xy2+2)$

9. $(2a2b2−3ab−1)$

10. $(x2n+xn+1)$

11. $4x2(3x2−4x+1)$

12. $4y3(5y5+7y3+10)$

13. $2a2b(a2b2−3ab+4)$

14. $x2y3(2xy2−4x2y+1)$

15. $(2x+3)(5x2−3)$

16. $(3x−1)(9x2+1)$

17. $xn(x4n−x2n+1)$

1. $(2x+3)(x2+1)$

2. $(2x−1)(3x2+2)$

3. $(x−1)(x2−3)$

4. $(2x+7)(x2−5)$

5. $(7y+5)(2y3−1)$

6. $(xn+1)(x3n+2)$

7. $(x−y)(x2+y2)$

8. $(x+3y)(3x2y2−1)$

9. $(a−4b)(ab−3)$

10. $(a2+b)(a2+b3)$

11. $(a−2b)(3x−5y)$

12. $xy2(x−y)(x2+y2)$

13. $a2b2(a2+b)(a+b2)$

1. $(x+8)(x−8)$

2. $(3+2y)(3−2y)$

3. $(x+9y)(x−9y)$

4. $(ab+2)(ab−2)$

5. $(ab+c)(ab−c)$

6. $(x2+8)(x2−8)$

7. $(3x+5)(x+5)$

8. $3(2y−3)$

9. $(3x+2)(x+8)$

10. $(x2+4)(x+2)(x−2)$

11. $(x2y2+1)(xy+1)(xy−1)$

12. $(x4+y4)(x2+y2)(x+y)(x−y)$

13. $(xn+yn)(xn−yn)$

14. $(x2n+y2n)(xn+yn)(xn−yn)$

15. $(x−3)(x2+3x+9)$

16. $(2y+3)(4y2−6y+9)$

17. $(x−y)(x2+xy+y2)$

18. $(2ab+1)(4a2b2−2ab+1)$

19. $(xy−5)(x2y2+5xy+25)$

20. $(2x+3)(x2+3x+9)$

21. $(x+1)(7x2+5x+1)$

22. $(xn−yn)(x2n+xnyn+y2n)$

23. $(a2+4)(a4−4a2+16)$

24. $(x+y)(x2−xy+y2)(x−y)(x2+xy+y2)$

25. $(xn+yn)(x2n−xnyn+y2n)×(xn−yn)(x2n+xnyn+y2n)$