Special products. Methods for factoring polynomials with integer coefficients. Highest common factor. Relatively prime polynomials. Lowest common multiple.

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Special products. The following are some products that occur frequently in mathematics and it is important to be familiar with them.

1. a(c + d) = ac + ad

2. (a + b)(a - b) = a² - b²

3. (x + a)(x + b) = x² + (a + b)x + ab

4. (ax + b)(cx + d) = acx² + (ad + bc)x + bd

5. (a + b)(c + d) = ac + bc + ad + bd

10. (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc

11. (a + b + c)³ = a³ + b³ + c³ + 3a²(b + c) + 3b²(a + c) + 3c²(a + b) + 6abc

In addition,

a³ - b³ = (a - b)(a² + ab + b²)

a⁴ - b⁴ = (a - b)(a³ + a²b + ab² + b³)

a⁵ - b⁵ = (a - b)(a⁴ + a³b + a²b² + ab³ + b⁴)

a⁶ - b⁶ = (a - b)(a⁵ + a⁴b + a³b² + a²b³ + ab⁴ + b⁵)

and, in general,

12. aⁿ - bⁿ = (a - b)(a^n-1 + a^n-2b + a^n-3b² + ... + ab^n-2+ b^n-1)

13. aⁿ - bⁿ = (a + b)(a^n-1 - a^n-2b + a^n-3b² - ... + ab^n-2- b^n-1)

where n is any positive integer (1, 2, 3, 4, ...).

Moreover,

a³ + b³ = (a + b)(a² - ab + b²)

a⁵ + b⁵ = (a + b)(a⁴ - a³b + a²b² - ab³ + b⁴)

a⁷ + b⁷ = (a + b)(a⁶ - a⁵b + a⁴b² - a³b³ + a²b⁴ - ab⁵ + b⁶)

and, in general,

14. aⁿ + bⁿ = (a + b)(a^n-1- a^n-2b + a^n-3b² - ... - ab^n-2+ b^n-1)

where n is any positive odd integer (1, 3, 5, 7, ...).

Methods for factoring polynomials with integer coefficients

Def. Factor. The factors of a given algebraic expression consist of two or more algebraic expressions which when multiplied together produce the given expression. For example, the algebraic expression x² + 2xy - 8y² can be written as the product (x + 4y)(x - 2y).

We shall consider here methods of decomposing a particular type of polynomial, namely one with integer coefficients, into products of polynomials of the same kind i.e. polynomials whose coefficients are also integers. A polynomial which cannot itself be further decomposed into factors whose coefficients are integers is said to be prime. A polynomial is said to be factored completely when it is expressed as a product of prime factors.

The process of factoring a polynomial is a trial and error procedure involving ingenuity and creativity. And there is no guarantee that a given polynomial can be factored. What follows is a list of methods that one should keep in mind when attempting to factor a polynomial. They represent possibilities to be considered. One of them may work.

Possible factoring methods.

1. Common monomial factor.

Type: ac + ad = a(c + d)

Examples.

8x³y + 4x² = 4x²(2xy + 1)

x³y + 4x² y - 5x²y³ = x²y(x + 4 - 5y²)

2. Difference of two squares.

Type: a² - b² = (a + b)(a - b)

Examples.

4x² - y² = (2x + y)(2x - y)

20a²x² - 45y² = 5(2ax + 3y)(2ax - 3y)

3. Trinomials that are perfect squares.

Type: a² + 2ab + b² = (a + b)²

a² - 2ab + b² = (a - b)²

Examples.

x² + 6x + 9 = (x + 3)²

16x² - 24xy² + 9y⁴ = (4x - 3y²)²

4. Trinomials of the form x² + qx + r.

Type: x² + (a + b)x + ab = (x + a)(x + b)

To factor x² + qx + r, we seek two numbers, a and b, such that ab = r and (a + b) = q. To find a and b, we examine the different pairs of factors of r in order to find a pair whose algebraic sum is q. Another method: Use the Quadratic formula to find the roots of the equation x² + qx + r = 0.

Examples.

x² - 5x + 4 = (x - 4)(x - 1) where a = -4, b = -1 so that their sum (a + b) = -5 and their product ab = 4

x² - 4x - 12 = (x - 6)(x + 2)

5. Trinomials of the form px² + qx + r.

Type: acx² + (ad + bc)x + bd = (ax + b)(cx + d)

The easiest way to factor this type is by using the Quadratic formula to find the roots of the equation px² + qx + r = 0.

Examples.

3x² - 5x - 2 = (x - 2)(3x + 1) where a = 1, c = 3, b = -2, d =1 found by trial and error

4x² - 8x -5 = (2x - 5)(2x + 1)

6. Sum or difference of two cubes.

Type: a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

Examples.

8x³ + 27y³ = (2x)³ + (3y)³

= (2x + 3y)[(2x)² - (2x)(3y) + (3y)²]

= (2x + 3y)(4x² - 6xy + 9y²)

27x³ - 64y³ = (3x - 4y)(9x² + 12xy + 16y²)

7. Grouping of terms.

Type: ac + bc + ad + bd = c(a + b) + d(a + b) = (a + b)(c + d)

Examples.

2ax - 4by + ay - 2by = 2x(a - 2b) + y(a - 2b) = (a - 2b)(2x + y)

3ax - ay - 3bx + by = a(3x - y) - b(3x - y) = (3x -y)(a - b)

8. Factors of aⁿ bⁿ.

Type: aⁿ - bⁿ = (a - b)(a^n-1 + a^n-2b + a^n-3b² + ... + ab^n-2+ b^n-1) where n is any positive integer (1, 2, 3, 4, ...).

aⁿ + bⁿ = (a + b)(a^n-1- a^n-2b + a^n-3b² - ... - ab^n-2+ b^n-1)where n is any positive odd integer (1, 3, 5, 7, ...).

Examples.

64 + y³ = 4³ + y³ = (4 + y)(4² - 4y + y²) = (4 + y)(16 - 4y + y²)

x⁷-1 = (x - 1)(x⁶ + x⁵ + x⁴ + x³ + x² + x + 1)

9. Adding and subtracting some perfect square.

Sometimes the addition and subtraction of a perfect square will change a polynomial into a difference of two squares.

Problem. Factor x⁴ + x²y² + y⁴.

Solution. Add and subtract x²y².

x⁴ + x²y² + y⁴ = x⁴ + x²y² + y⁴ + x²y² - x²y²

= x⁴ + 2x²y² + y⁴ - x²y²

= (x² + y²)² - x²y²

= [(x² + y²) + xy][(x² + y²) - xy]

= (x² + xy + y²)(x² - xy + y²)

Def. Highest common factor (H. C. F.) of two or more polynomials. The polynomial of highest degree and largest numerical coefficients (apart from trivial changes in sign) which is a factor of all the given polynomials.

Syn. Greatest common divisor

To find the highest common factor of several polynomials first write each polynomial as a product of prime factors. Then note those factors common to all the polynomials and take each common factor to the lowest power to which it occurs in all polynomials.

Problem. Find the H. C. F. of

2²3³(3x + y)⁴(x + 1)², 3³x(3x + y)²(x + 1)³, 2³3²(3x + y)³(x + 1)

Solution. The common factors are 3, (3x + y) and (x + 1). The H. C. F. is 3²(3x + y)²(x + 1).

Def. Relatively prime polynomials. Two polynomials are relatively prime if their H. C. F. is 1, that is, they have no common factors other than 1.

Def. Lowest common multiple (L. C. M.) of two or more polynomials. The polynomial of lowest degree and smallest numerical coefficients (apart from trivial changes in sign) which contains each of the given polynomials as a factor.

To find the lowest common multiple of several polynomials first write each polynomial as a product of prime factors and then make a list of all the prime factors that occur in the various polynomials. The L. C. M. is the product obtained by multiplying together all of the prime factors in the list, taking each factor to the highest power to which it occurs in any of the given polynomials.

Problem. Find the L. C. M. of

2²3³(3x + y)⁴(x + 1)², 3³x(3x + y)²(x + 1)³, 2³3²(3x + y)³(x + 1)

Solution. The list of factors is 2, 3, x, (3x + y) and (x + 1). The L. C. M. is 2³3³x(3x + y)⁴(x + 1)³.

References

Hawks, Luby, Touton. Second-Year Algebra

Murray R. Spiegel. College Algebra

Raymond W. Brink. A First Year of College Mathematics