Factoring Polynomials

Factoring polynomials means breaking down a polynomial (with two, three, or more terms) into simpler expressions or factors that, when multiplied together, give back the original polynomial. 

For example, the polynomial x² + 3x + 2 can be factored as (x + 1)(x + 2)

Factoring is useful for simplifying polynomials and for finding the zeros of polynomial functions by setting each factor to zero. For the polynomial x² + 3x + 2, the zeros are x = -1 and x = -2. Factoring also helps us to analyze the polynomial functions.

Here are the different ways to factor polynomials:

Greatest Common Factor (GCF) Method

This method is the reverse of the distributive law, which states that a(b + c) = ab + ac

For factoring the polynomial ab + ac, we just take out the greatest common factor:

ab + ac = a(b + c), here a is the greatest common factor.

Now, let us factor the polynomial 6x² + 12x 

Finding the GCF of the Terms

6x² + 12x is a binomial (having 2 terms). 

Here, both terms have coefficients that are divisible by 6, and both have at least one x variable, so the GCF, in this case, is 6x.

Thus, the GCF of 6x² and 12x is 6x

Factoring Out the GCF

6x² + 12x

= 6x ⋅ x + 6x ⋅ 2

= 6x(x + 2)

Thus, 6x and (x + 2) are the two factors of the polynomial 6x² + 12x

Middle-Term Method

For polynomials with 3 terms (trinomials, quadratic, or 2^nd-degree polynomials) of the form ax² + bx + c (here, a ≠ 0), first, we need to find the two numbers that multiply to ac and then add up to b (middle term). We then rewrite the trinomial by breaking down the middle term as ax² + (m + n)x + c and group the terms to factor. 

Here, m + n = b and mn = ac

When a = 1

Let us factor the polynomial x² – 3x – 10 …..(i)

Comparing With the Standard Form

As we know, the standard form of a trinomial is ax² + bx + c …..(ii)

Comparing the expressions (i) and (ii), we get

a = 1, b = -3, and c = -10

Thus, ac = (1)(-10) = -10 

Finding Two Numbers

Since ac = -10, we need two numbers that multiply to give -10 and add up to -3

Here, -10 = (-5)(2), and b = -3 = -5 + 2

Writing the Trinomial

Now, x² – 3x – 10

= x² + (-5 + 2)x + (-5)(2)

= x² – 5x + 2x – 10

= x(x – 5) + 2(x – 5)

= (x – 5)(x + 2)

Thus, x² – 3x – 10 is factored as (x – 5)(x + 2)

When a ≠ 1

Let us factor the polynomial 2x² + 9x + 10 …..(i)

Comparing With the Standard Form

As we know, the standard form of a trinomial is ax² + bx + c …..(ii)

Comparing the expressions (i) and (ii), we get

a = 2, b = 9, and c = 10

Thus, ac = (2)(10) = 20 

Finding Two Numbers

Since 20 = (5)(4) and b = 9 = 5 + 4

Here, the numbers are 5 and 4, which multiply to get 20 and add up to 9.

Writing the Trinomial

Now, 2x² + 9x + 10

= 2x² + (5 + 4)x + 10

= 2x² + 5x + 4x + 10

= x(2x + 5) + 2(2x + 5)

= (2x + 5)(x + 2)

Thus, 2x² + 9x + 10 = (2x + 5)(x + 2)

Grouping Method

This method is useful for polynomials with four or more terms (known as cubic or 3^rd-degree polynomials). To factor a polynomial by grouping, the terms of the given polynomial are grouped in pairs to find the zeroes. The GCF is then factored out from each group.

Let us factor the polynomial x³ + 3x² + 2x + 6

Grouping the Terms in Pairs

x³ + 3x² + 2x + 6

= (x³ + 3x²) + (2x + 6)

Factoring Out the GCF From Each Group

= x²(x + 3) + 2(x + 3)

Combining the Factored Groups

= (x + 3)(x² + 2)

Thus, (x + 3)(x² + 2) is the factored form of the polynomial x³ + 3x² + 2x + 6

Here is a summary of the different methods of factoring polynomials.

The ‘Box or ‘Grid’ method for factoring polynomials is an alternative to the grouping method, especially when the leading coefficient is a ≠ 1 or a ≠ -1

Let us factor the trinomial 2x² + 7x + 6 …..(i)

Comparing With the Standard Form

As we know, the standard form of a trinomial is ax² + bx + c …..(ii)

Comparing the expressions (i) and (ii), we get

a = 2, b = 7, and c = 6

Thus, ac = (2)(6) = 12 

Finding Two Numbers

Since 12 = (3)(4) and b = 7 = 3 + 4

Here, the numbers are 3 and 4, which multiply to get 12 and add up to 7.

Forming the Box

Here, we place the terms in their corresponding boxes:

• The first term is placed in the upper left box.
• The constant is placed in the lower right box.
• The numbers (found in step 2) with a variable x are then placed in the remaining empty boxes, irrespective of their orders. 

Finding the greatest common factor in each row and column

Here,

GCF of the first row is GCF(2x², 4x) = 2x

GCF of the second row is GCF(3x, 6) = 3

GCF of the first column is GCF (2x², 3x) = x

GCF of the second column is GCF (4x, 6) = 2

Now, placing them on the side and top of the box, we get

Writing the Polynomial in Factored Form

The first factor is found by adding the terms in the left column, while the second factor is determined by adding the terms in the top row.

Thus, the answer is (2x + 3)(x + 2)

Special Forms – By Using Identities

Some special forms of polynomials can be easily factored using identities.

Factoring Difference of Squares

A polynomial of the form a² – b² can be factored as: 

• a² – b² = (a + b)(a – b)

Let us factorise the polynomial 9x² – 49y² …..(i)

Comparing

Now, comparing (i) with the identity a² – b², we get

a² = 9x² and b² = 49y²

⇒ a = 3x and b = 7y

Factoring

Here, 9x² – 49y²

= (3x)² – (7y)² 

= (3x + 7y)(3x – 7y)

Thus, 9x² – 49y² = (3x + 7y)(3x – 7y)

Factoring Perfect Square Trinomials

A perfect square trinomial can be factored into the square of a binomial: 

• a² + 2ab + b² = (a + b)² 
• a² – 2ab + b² = (a – b)² 

Let us factorise x² + 6x + 9 …..(i)

Comparing

Comparing equation (i) with the form a² + 2ab + b², we get

a² = x² and b² = 9

⇒ a = x and b = 3 

Factoring

Here, x² + 6x + 9

= (x)² + 2(x)(3) + (3)² 

= (x + 3)² 

Thus, x² + 6x + 9 = (x + 3)² = (x + 3)(x + 3)

Factoring the Sum and Difference of Cubes

Polynomials of the form a³ + b³ and a³ – b³ can be easily factored using the given identities:

• a³ + b³ = (a + b)(a² – ab + b²)
• a³ – b³ = (a – b)(a² + ab + b²)

Let us factorise x³ + 27y³ …..(i)

Comparing

Comparing equation (i) with the form a³ + b³, we get

a³ = x³ and b³ = 27y³ 

⇒ a = x and b = 3y 

Factoring

Here, x³ + 27y³ 

= (x)³ + (3y)³ 

= (x + 3y)[(x)² + (x)(3y) + (3y)²]

= (x + 3y)(x² + 3xy + 9y²)

Thus, x³ + 27y³ = (x + 3y)(x² + 3xy + 9y²)