# Factoring Cubic Polynomials (+Questions)

Factoring polynomials is necessary for solving many types of math problems. Sometimes it becomes challenging when we encounter a cubic polynomial. One way is to find the roots by applying the cubic formula, but it is too complex to remember and use. However, there are alternative methods for factoring these polynomials.

The three methods we use for factoring a cubic polynomial are splitting terms using the ad-method, finding a factor by applying the rational root theorem, and cubic formulas for sum, difference, etc.

## Irreducible Polynomials

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

## The Standard Form of Cubic Polynomial

As we are focused on the cubic polynomial, we need its standard form for our discussion.

### ✩ Standard form of the Cubic Polynomial

p(x) = ax3 + bx2 + cx + da ≠ 0

## A. The “ad” Method for Linear Factors

This method is named ” ad “ because coefficients a and d play a central role in the factorization process. In this method, we split the coefficients b and c of the standard cubic polynomial as:

ax3 + (X1 + X2 + X3)x2 + (Y1 + Y2 + Y3)x + d

We use the following steps to achieve this result.

### Step 1. Split the Coefficient b

We first find X1, X2, X3 using the following conditions:

1. Each of X1, X2, X3 is a factor of ad
2. X1.X2.X3 = a2d
3. X1 + X2 + X3 = b

### Step 2. Split the Coefficient c

Now calculate Y1, Y2, Y3 using the formulas:

Finally, we verify that:

2. Y1 + Y2 + Y3 = c

If the verification fails, we go back to step 1 and find another triplet X1, X2, X3.

Once we have X1, X2, X3 and Y1, Y2, Y3, rest of the factorization is simple algebra.

If steps appear difficult, do not worry, as, in practice, they are pretty simple. We have to find only X1, X2, X3 the rest falls in place automatically!

All the following conditions must be satisfied for successful factorization of the cubic polynomial:

1. X1 + X2 + X3 = b
2. X1.X2.X3 = a2d
3. Y1 + Y2 + Y3 = c
5. Each of Xi and Yi is a factor of ad
6. X1.Y1 = X2.Y2 = X3.Y3 = ad

In practice the conditions 1, 2 and 5 are used to find X1, X2, X3. Calculation of Y1, Y2, Y3 is straightforward using the equations in the condition 6.

### Example 1

Factorize 2y3 + 23y2 + 8y − 33

#### Step 1. Split the Coefficient b

Comparing this polynomial with the standard form, we get the coefficients as:

a = 2, b = 23, c = 8 and d = − 33

We split the term b as a sum of three numbers.

b = X1 + X2 + X3

To do so we first calculate:

• a2d = 22( − 33) = − 132
• ad = 2( − 33) = − 66

Now we find factors X1, X2 and X3 of ad = − 66 such that:

1. X1.X2.X3 = a2d = − 132
2. Their sum is equal to the coefficient b = 23

Prime factors of a2d = − 132 are:
−132 = − (2 × 2 × 3 × 11)

We rewrite it as a product of three numbers:
= 22 × 3 × ( − 2)

This gives us: X1 = 22, X2 = 3, X3 = − 2

The sum of 22, 3 and −2 is 23 = b. Also each number is a factor of -66.

#### Step 2. Split the Coefficient c

Just like the coefficient b, we have to split the coefficient c as sum of three numbers (Y1, Y2, Y3).

We calculate Y1, Y2, Y3 using the formulas:

Let us check if these sum up to the coefficient c:

Y1 + Y2 + Y3
= − 3 − 22 + 33 = − 25 + 33
= 8 = c

Also their product should be equal to ad2 = 2( − 33)2 = 2178
Y1.Y2.Y3 = ( − 3)( − 22)(33) = 2178

Both conditions are satisfied!

Now rest of the factorization is a simple algebraic grouping and taking out common factors.

#### Step 3. Grouping

We now split the coefficients b and c of the cubic polynomial using the results of the previous steps.

2y3 + 23y2 + 8y − 33

= 2y3 + (22 + 3 − 2)y2 + ( − 3 − 22 + 33)y − 33

= 2y3 + 22y2 + 3y2 − 2y2 − 3y − 22y + 33y − 33

Grouping the terms:

= (2y3 − 2y2) + (3y2 − 3y) + (22y2 − 22y) + (33y − 33)

Taking out the common factors:

= 2y2(y − 1) + 3y(y − 1) + 22y(y − 1) + 33(y − 1)

Taking out (y − 1) as a common factor:

= (y − 1)[2y2 + 3y + 22y + 33]

Again grouping terms and taking out the common factors in the second bracket:

= (y − 1)[2y(y + 11) + 3(y + 11)]

Taking out (y + 11) as a common factor:

= (y − 1)(y + 11)(2y + 3)

Solved!

### Does this method always work?

• The ad method works if three linear factors exist for the cubic polynomial
• If they do not exist, you will not be able to find X1, X2, X3 in the first step
• In most of the questions, it is relatively easy to see that X1, X2, X3 do not exist
• If X1, X2 and X3 do not exist, one of the factors is quadratic or factorization is not possible

The proof and details of this method are in the article published on the Northern Michigan University website.

Now let us see an example of a polynomial where this method does not work.

### Example 2

Factorize p(x) = 3x3 + 5x2 − x + 2

#### Step 1. Split the Coefficient b

The coefficients p(x) are a = 3, b = 5, c = − 1, d = 2

We calculate a2d = 32(2) = 18 and ad = 3(2) = 6

We have to find X1, X2 and X3 such that:

1. Each is a factor of ad = 6
2. Their sum is equal to the coefficient b = 5
3. Their product is equal to a2d = 32(2) = 18

Since each of X1, X2, X3 is also a factor of 6, only possible triplet is 6, 3, 1. The possible combinations of this triplet are:

• 6, 3, 1
• 6, − 3, − 1
• −6, − 3, 1
• −6, 3, − 1

The product of all the above triplets results in 18, yet, none of them sum up to 5. So this method will not work in this case.

What can we do now? We factorize this polynomial in the next section.

## B. Using the Rational Root Theorem

According to the Rational Root Theorem, the roots of a polynomial are among the following numbers:

In our case, this means roots of the polynomial belong to the set:

Root Set =

As the name suggests, this theorem works if the roots of the polynomial are rational.

We know from the factor theorem that if the number “r” is a root of the polynomial p(x), then one of its factors is ( x – r ).

Let us now try this method on the polynomial we failed to factor in the previous example.

### Example 3

Factorize p(x) = 3x3 + 5x2 − x + 2

Coefficients of the polynomial are:

a = 3, b = 5, c = − 1, d = 2

According to the rational root theorem, the roots lie among:

Let us evaluate these:

• p(1) = 3.13 + 5.12 − 1 + 2 = 9
• p(1) ≠ 0
• So 1 is not a root.
2. Let us try -1:
• p( − 1) = 3( − 1)3 + 5( − 1)2 − ( − 1) + 2 = 5
• p( − 1) ≠ 0
• So -1 is not a root
3. As 5x2 is positive for all values of x, we need a negative term from 3x3. For that, we need to try negative numbers. Therefore let us try -2, instead of 2:
• p( − 2) = 3( − 2)3 + 5( − 2)2 − ( − 2) + 2 = − 24 + 20 + 2 + 2 = 0p( − 2) = 0
• So -2 is a root!

Therefore (x − ( − 2)) = (x + 2) is a factor of p(x).

We can either divide p(x) by (x + 2) or cleverly split its terms to get (x + 2) as common factor. We take the second approach.

p(x) = 3x3 + 5x2 − x + 2

= 3x3 + 6x2 − x2 − 2x + x + 2

If the previous step is perplexing, merely divide the polynomial and factor the quotient.

Taking out the common factors we get:

= 3x2(x + 2) − x(x + 2) + (x + 2)

Taking out (x + 2) as common:

= (x + 2)(3x2 − x + 1)

We cannot factor the quadratic polynomial in the second bracket as its discriminant is negative.

Discriminant = B2 − 4AC = ( − 1)2 − 4(3)(1) = 1 − 12 = − 11

So factors of p(x) are (x + 2)(3x2 − x + 1)

### The Factor Theorem

The factor theorem tells us that for any polynomial p(x) if p(a) = 0, (x − a) is a factor of p(x).

So if we find a root of the polynomial p(x) we have a factor for p(x).

Example: p(x) = x3 − 3x2 + x + 1

Putting x = 1 in the expression we get:

p(1) = 13 − 3(1)2 + 1 + 1

= 1 − 3 + 2 = 3 − 3 = 0

According to factor theorem, (x − 1) is a factor of the polynomial p(x).

You can check that p(x) = (x − 1)(x2 − 2x − 1)

There is a formula for finding roots of a cubic polynomial, though it is very complex. Read the following articles if it interests you:

## C. Using the Cubic Formulas

For many polynomials, using formulas makes factorization easy. Remembering the formulas is essential to recognize patterns resembling them in a polynomial. We do not want to miss an opportunity for their application.

In some polynomials, you may find only a few terms of the formula are present. We transform the polynomial in such a case, making it suitable for formula application.

### ✩ Important Formulas

1. Sum and Difference of Cubes
• a3 + b3 = (a + b)(a2 − ab + b2)
• a3 − b3 = (a − b)(a2 + ab + b2)
2. Perfect Cubes
• (a + b)3 = a3 + b3 + 3a2b + 3ab2
• (a − b)3 = a3 − b3 − 3a2b + 3ab2
3. Difference of Squares ✩
• a2 − b2 = (a − b)(a + b)
• This is one of the most important formula. It is widely used for factoring polynomials.
4. Perfect Squares
• (a + b)2 = a2 + 2ab + b2
• (a − b)2 = a2 − 2ab + b2

### Example 4

Factorize: p(w) = w3 + 9w2 + 27w + 19

The formula for the expansion of the perfect cube is:

(a + b)3 = a3 + b3 + 3a2b + 3ab2

Observe that:

• 9w2 = 3(w2 × 3) = 3a2ba = w, b = 3
• 27w = 3(w × 32) = 3ab2a = w, b = 3

These are direct terms from the expansion of (w + 3)3:

(w + 3)3 = (w3 + 9w2 + 27w + 33)

We need only 33 = 27 in the polynomial to apply the formula.

So we split 19 as 27 − 8. Now we get a perfect cube:

p(w) = (w3 + 9w2 + 27w + 27) − 8

= (w3 + 3(w2 × 3) + 3(w × 32) + 33) − 8

= (w + 3)3 − 8

As 8 = 23, we can directly use the formula:

a3 − b3 = (a − b)(a2 + ab + b2).

p(w) = (w + 3)3 − 23

= [(w + 3) − 2][(w + 3)2 + (w + 3) × 2 + 22]

= (w + 1)(w2 + 6w + 9 + 2w + 6 + 4)

= (w + 1)(w2 + 8w + 19)

The second bracket has a quadratic polynomial with coefficients:

A = 1, B = 8, C = 19

Its discriminant D = B2 − 4AC = 64 − 76 = − 12

It is impossible to factorize this quadratic polynomial. Its discriminant is negative.

So our factors are:
(w + 1)(w2 + 8w + 19)

## Practice Questions

### 1 Question

Factorize p(x) = x3 + 6x2 + 11x + 6

Can you use the Rational Root Theorem to find a root/factor of p(x)?

### 2 Question

Factorize p(x) = x6 − 1

Can you apply the formula for a3 − b3?

### 3 Question

Factorize p(x) = x3 − 7x2 + 7x + 15

Can you split the coefficient of x2 in a triplet using the ad-method?

### 4 Question

Factorize: p(x) = x3 + 5x2 + 3x − 9

What is the value of expression for x = 1? Can you guess a factor now?

### 5 Question

Factorize p(x) = 2x3 + 3x2 + 2x + 1

What is the value of expression for x = − 1? Can you guess a factor?

The coefficients of this polynomial are:

a = 1, b = 6, c = 11, d = 6

According to the Rational Root Theorem, roots of the polynomial are among the following numbers:

p(x) will always evaluate to a value greater than 0 for any positive x.

So we try negative numbers only. Let us try start with -1:

p( − 1) = ( − 1)3 + 6( − 1)2 + 11( − 1) + 6 = − 1 + 6 − 11 + 6 = 0

We have our factor as: (x − ( − 1)) = (x + 1).

p(x) = x3 + 6x2 + 11x + 6

We could have divided p(x) by (x + 1), but the terms are easily split for this polynomial. While splitting terms, remember our aim is to get (x + 1) as a common factor.

= x3 + x2 + 5x2 + 5x + 6x + 6

Taking out the common terms:

= x2(x + 1) + 5x(x + 1) + 6(x + 1)

Taking out (x + 1) as common:

= (x + 1)[x2 + 5x + 6]

The quadratic polynomial in the second bracket is factored by splitting the middle term using the “ac” method.

= (x + 1)[x2 + 2x + 3x + 6]

Taking out the common terms:

= (x + 1)[x(x + 2) + 3(x + 2)]

Taking out (x + 2) as common:

= (x + 1)(x + 2)(x + 3)

That’s all!

Clearly, this question is apt for applying the formula for the difference of cubes as x6 − 1 = (x2)3 − 13.

p(x) = x6 − 1

= (x2)3 − 13

Applying the formula: a3 − b3 = (a − b)(a2 + ab + b2)

= (x2 − 1)(x4 + x2 + 1)

The first bracket can be written as difference of squares:

= (x2 − 12)(x4 + x2 + 1)

We apply the formula: a2 − b2 = (a − b)(a + b)

= (x − 1)(x + 1)(x4 + x2 + 1)

We can complete the square by adding and subtracting x2 to the terms of the last bracket:

= (x − 1)(x + 1)(x4 + 2x2 + 12 − x2)

= (x − 1)(x + 1)((x2)2 + 2(x2)(1) + 12 − x2)

= (x − 1)(x + 1)((x2 + 1)2 − x2)

Applying the difference of squares ( a2 − b2 = (a + b)(a − b) ) formula to the terms of the last bracket:

= (x − 1)(x + 1)(x2 + 1 − x)(x2 + 1 + x)

= (x − 1)(x + 1)(x2 − x + 1)(x2 + x + 1)

That’s all!

Standard form of cubic polynomial is ax3 + bx2 + cx + d

Coefficients of p(x) : a = 1, b = − 7, c = 7, d = 15

We calculate ad = 15, a2d = 15

For ad-method, our triplet X1, X2, X3 should be such that:

1. Each Xi is a factor of ad = 15

2. X1.X2.X3 = a2d = 15

3. X1 + X2 + X3 = b = − 7

Prime factors of a2d = 15 = 3 × 5

If we write a2d = 15 = 1 × ( − 3) × ( − 5), we get: X1 = 1, X2 = − 3, X3 = − 5.

X1 + X2 + X3 = 1 − 3 − 5 = − 7

The triplet satisfies the three conditions above.

We can calculate triplet Y1, Y2, Y3 using the formula:

Let us verify the conditions for suitability of Y1, Y2, Y3:

1. Y1.Y2.Y3 = 15( − 5)( − 3) = 225 = ad2

2. Y1 + Y2 + Y3 = 15 − 5 − 3 = 7 = c

Both the conditions are satisfied. So we now have a valid triplet splitting the coefficient c.

We split the coefficients using the triplets X1, X2, X3 and Y1, Y2, Y3:

p(x) = x3 − 7x2 + 7x + 15

= x3 + (1 − 3 − 5)x2 + (15 − 5 − 3)x + 15

= x3 + x2 − 3x2 − 5x2 + 15x − 5x − 3x + 15

Rearranging the terms and grouping them:

= (x3 + x2) + ( − 3x2 − 3x) + ( − 5x2 − 5x) + (15x + 15)

Taking out the common factors:

= x2(x + 1) − 3x(x + 1) − 5x(x + 1) + 15(x + 1)

Taking out (x + 1) as a common factor:

= (x + 1)[x2 − 3x − 5x + 15]

= (x + 1)[x(x − 3) − 5(x − 3)]

= (x + 1)(x − 3)(x − 5)

The coefficients of the p(x) : a = 1, b = 5, c = 3, d = − 9

Observe the sum of coefficients a + b + c + d = 0. The polynomial evaluates to zero if we replace x by 1 (p(1) = 0). Therefore (x − 1) is a factor of p(x).

Rational root theorem also gives the root, but it was easier to guess for this polynomial.

As the next step, we can divide p(x) by (x − 1) and factorize the resulting quotient, but we choose to split the terms.

We force our way to get (x − 1) as a factor for each pair of reducing exponents of x.

If we pair −x2 with x3 we get (x − 1) as a factor. So we add ( − x2 + x2) to p(x):

p(x) = x3 − x2 + x2 + 5x2 + 3x − 9

= x2(x − 1) + 6x2 + 3x − 9

If we pair −6x with 6x2 we get (x − 1) as a factor. So we add ( − 6x + 6x) to p(x):

= x2(x − 1) + 6x2 − 6x + 6x + 3x − 9

= x2(x − 1) + 6x(x − 1) + 9x − 9

= x2(x − 1) + 6x(x − 1) + 9(x − 1)

Taking out (x − 1) as a common factor:

= (x − 1)(x2 + 6x + 9)

It worked out beautifully! It could not have gone wrong as (x − 1) is a known factor.

Rewriting the terms in the second bracket:

= (x − 1)(x2 + 2(x)(3) + 32)

The terms in the second bracket are expanded form of a perfect square:

(a + b)2 = a2 + 2ab + b2a = x, b = 3.

Therefore p(x) = (x − 1)(x + 3)2

Coefficients of p(x) : a = 2, b = 3, c = 2, d = 1.

From the Rational Root Theorem roots of p(x) are among the following:

As all the coefficients are positive, p(x) > 0 for any positive x. Therefore root of p(x) is a negative number.

Let us try x = − 1:

p( − 1) = 2( − 1)3 + 3( − 1)2 + 2( − 1) + 1

= − 2 + 3 − 2 + 1 = 0

x = − 1 is a root of the polynomial p(x).

From the factor theorem, we know that x + 1 is a factor of p(x).

If we pair 2x2 we 2x3 we get (x + 1) as a factor. Therefore we write 3x2 = 2x2 + x2

p(x) = 2x3 + 3x2 + 2x + 1

= 2x3 + 2x2 + x2 + 2x + 1

= 2x2(x + 1) + x2 + 2x + 1

Observe that x2 + 2x + 1 is a perfect square of (x + 1).

= 2x2(x + 1) + (x + 1)2

Taking out the common factor (x + 1):

= (x + 1)(2x2 + x + 1)

In the second bracket, we have a quadratic polynomial but, it is irreducible. Can you prove it?

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