# Factoring Polynomials (Methods + Questions)

Factoring polynomials can be challenging, particularly for degrees 3 (or more) and containing multiple variables. There are many ways to factorize them. We choose a method depending on the polynomial.

## List of Factorization Methods

The polynomials can be factorized by completing the square, splitting terms, finding roots (rational root theorem), ad-method, and using quadratic/cubic identities (sum, difference, etc).

1. Difference of squares
2. Completing the square
3. Splitting terms
4. Rational root theorem
6. Formulas: perfect cube, sum/difference of cubes
7. Splitting the middle term (quadratic polynomials)

### Irreducible Polynomials

Polynomials like 2x + 1 or 3x2 − x + 1 cannot be factored. These are irreducible polynomials. All factorization methods aim to represent a polynomial as a product of two (or more) lower degree polynomials. The factorization is complete when the resulting factors are irreducible.

## 1. Difference of Squares

The formula a2 − b2 = (a − b)(a + b) provides direct factorization if the difference of squares appears in the expression. It is a commonly used formula, and you should certainly remember it. Let’s factorize a polynomial using it.

### 1 Example

Factorize

#### Solution

Using the formula for difference of squares a2 − b2 = (a − b)(a + b) we get:

= x2 − 9

= x2 − 32

= (x − 3)(x + 3)

You can also factor polynomials of degrees more than 2 using this formula. We have to recognize the a2 − b2 pattern. The degree must be even. Let us see an example.

### 2 Example

Factorize y4 − 25

#### Solution

= (y2)2 − 25

= (y2)2 − 52

= (y2 − 5)(y2 + 5)

## 2. Completing the Square

We use the following formulas to make a perfect square.

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

We look for the terms on the left-hand side of the formula in the given polynomial. These terms are usually not apparent (or missing), for example:

1. We may get a4 instead of a2
2. 2ab part may not be exact
3. b2 may be missing

Our task is to identify the pattern or make changes in the polynomial to match the formula. Let us see an example.

### 3 Example

Factorize y2 + 12y + 36

#### Solution

We will try to apply the perfect square formula (a + b)2 = a2 + 2ab + b2

Rewriting y2 + 12y + 36:

= y2 + 12y + 62

Looking at the squared terms (first and last), we get: a = y, b = 6

Rewriting the expression to see if these values satisfy its middle term:

= y2 + 2(y)(6) + 62

They do! So we have a perfect square:

= (y + 6)2

We have two variables in the following polynomial. We will factorize this polynomial by making it a perfect square.

### 4 Example

Factorize p(x) = 4x2 − 12xy + 5y2

#### Solution

Rewriting 4x2 − 12xy to fit the perfect square formula (a − b)2 = a2 − 2ab + b2:

4x2 − 12xy = (2x)2 − 2(2x)(3y)

a = 2x, b = 3y but b2 = (3y)2 is missing (Compare with the formula)

Add and subtract b2 = (3y)2. This does not change the original expression ( we are adding 0 ! ), however we will be able to complete the square.

4x2 − 12xy = (2x)2 − 2(2x)(3y) + (3y)2 − (3y)2

4x2 − 12xy = (2x − 3y)2 − (3y)2

Using above in p(x):

= 4x2 − 12xy + 5y2

= (2x − 3y)2 − (3y)2 + 5y2

= (2x − 3y)2 − 9y2 + 5y2

= (2x − 3y)2 − 4y2

Rewriting the above expression we get difference of squares:

= (2x − 3y)2 − (2y)2

= (2x − 3y − 2y)(2x − 3y + 2y)

= (2x − 5y)(2x − y)

## 3. Splitting Terms

We often factorize quadratic polynomials by splitting the middle term. For them, the method is well defined and easy to use. For the higher degree polynomials, we use the coefficients (and degrees) of neighboring terms to guide us. You will get the hang of the “splitting terms” technique with practice.

### 5 Example

Factorize 2x3 − 9x2 − 8x + 15

#### Solution

We have 15 at the end, so we try to split −8x as −15x + 7x.

= 2x3 − 9x2 + 7x − 15x + 15

= 2x3 − 9x2 + 7x − 15(x − 1)

We will now split −9x2 = − 2x2 − 7x2. Why?

a. 7x2 and −7x have (x − 1) as factor

b. 2x3 and −2x2 have (x − 1) as factor

This works great!

= 2x3 − 2x2 − 7x2 + 7x − 15(x − 1)

= 2x2(x − 1) − 7x(x − 1) − 15(x − 1)

= (x − 1)(2x2 − 7x − 15)

Second bracket contains a quadratic expression. It is easy to split the middle term.

= (x − 1)(2x2 − 10x + 3x − 15)

= (x − 1)(2x(x − 5) + 3(x − 5))

= (x − 1)(x − 5)(2x + 3)

## Questions

### 1 Question

Factorize 4a4 + 5a2 + 1

Can you split the middle term?

### 2 Question

Factorize x4 + 324

Can you try to complete the square?

### 3 Question

Factorize

Can you open the bracket and try?

### 4 Question

Factorize a4 + 4a2 − 5

What if we write 4a2 = 5a2 − a2 ?

### 5 Question

Factorize a4 + a2b2 + b4

Can you complete the square?

Splitting the middle term 5a2 as 4a2 + a2

= 4a4 + 4a2 + a2 + 1

= 4a2(a2 + 1) + a2 + 1

= (a2 + 1)(4a2 + 1)

Observe that 324 = 182

= x4 + 182

Adding and subtracting 2 × 18 × x2

= (x2)2 + 182 + (2 × 18 × x2) − (2 × 18 × x2)

= (x2 + 18)2 − 36x2

36x2 can be written as (6x)2. We can then use the formula a2 − b2 = (a − b)(a + b).

= (x2 + 18)2 − (6x)2

= (x2 + 18 − 6x)(x2 + 18 + 6x)

= (x2 − 6x + 18)(x2 + 6x + 18)

Let us open the bracket:

Applying the formula on the first bracket.

= a4 + 4a2 − 5

We split the term 4a2 as 5a2 − a2

= a4 + 5a2 − a2 − 5

= a4 − a2 + 5a2 − 5

= a2(a2 − 1) + 5(a2 − 1)

= (a2 − 1)(a2 + 5)

We can further factorize using the formula a2 − b2 = (a − b)(a + b) on first bracket:

= (a2 − 12)(a2 + 5)

= (a − 1)(a + 1)(a2 + 5)

= a4 + a2b2 + b4

We can complete the square by adding and subtracting a2b2:

= a4 + 2a2b2 − a2b2 + b4

= a4 + 2a2b2 + b4 − a2b2

= (a2)2 + 2a2b2 + (b2)2 − a2b2

= (a2 + b2)2 − (ab)2

Expression above is difference of squares. Using the formula a2 − b2 = (a − b)(a + b)

= [(a2 + b2) − (ab)][(a2 + b2) + (ab)]

= [a2 + b2 − ab][a2 + b2 + ab]

= [a2 − ab + b2][a2 + ab + b2]

## Related

Factoring a Quadratic in 4 Easy Steps (+ Questions) ➤

Factoring Cubic Polynomials (+ Questions) ➤

Finding Roots of Quadratic (Equation with Examples, Graphs) ➤