# Factoring Polynomials : Learn Definition, Types, Properties using Solved Examples!

Factoring polynomials is similar to finding the factors of a number. We divide the number by another number and write it as a product of its factors. We do the same with polynomials. There are many ways to do so. Factoring Polynomials helps us to solve polynomial algebraic equations. Polynomials are algebraic expressions that include real numbers and variables. They are one of the important introductory topics in algebra. For an expression to be a polynomial, the variables in the expression must have whole-number powers. In this article, we will learn the process of factoring polynomials, their importance, use of theorems that aid the process, and different ways to factor polynomials with solved examples.

## Factoring Polynomials

A polynomial can be represented as \(a_x^n + a_x^n-1..+ a_0 = 0\)

Factoring polynomials is one of the important steps in finding out the solution of the polynomial.

- The solution of a zero polynomial or the zeros of a polynomial are the values of the variables for which the polynomial gives us the answer as zero.
- A single polynomial can have multiple solutions or zeros.
- While finding the zeros of a polynomial one must remember that the number of zeros of a polynomial is equal to or less than the degree of polynomial.
- The number of linear factors of a polynomial will always be equal to the degree of the polynomial.
- Linear Factors are nothing but algebraic expressions having degree one.
- Hence, a linear algebra polynomial will have one factor, a quadratic polynomial will have two factors, a cubic polynomial will have three factors and so on.

Let’s see some examples.

\({x^2} – 25 = ({x + 5})({x – 5})\)

\(4{x^2} + 10x – 6 = 2( {2x – 1} )( {x + 3} )\)

## Stepwise Method in Factoring Polynomials

Here are the steps to factor polynomials:

- For all polynomials, the first factor out the greatest common factor (GCF).
- For a binomial, check to see if it is any of the following:

- Difference of squares: \(x^2 – y^2 = (x + y) (x – y)\)
- Difference of cubes: \(x^3 – y^3 = (x – y) (x^2 + xy + y^2)\)
- Sum of cubes: \(x^3 + y^3 = (x + y) (x^2 – xy + y^2)\)

- For a trinomial, check to see whether it is either of the following forms:

- \(x^2 + bx + c\): If so, find two integers whose product is c and whose sum is b. For example,
- \(x^2 + 8x + 12 = (x + 2)(x + 6)\). Since (2)(6) = 12 and 2 + 6 = 8
- \(ax^2 + bx + c\): If so, find two binomials so that
- The product of the first terms = \(ax^2\)
- The product of the last terms = \(c\)
- The sum of outer and inner products = \(bx\)

Learn about Zeros of a Cubic Polynomial

## Methods of Factoring Polynomials

Factoring is the process by which we go about determining what we multiplied to get the given quantity. There are various methods to do so. Let’s see them one by one.

Example: \({x^2} – 16 = ( {x + 4} )( {x – 4} )\)

\({x^4} – 16 = ( {{x^2} + 4} )( {{x^2} – 4} )\)

\({x^4} – 16 = ( {{x^2} + 4} )( {x + 2} )( {x – 2} )\)

### Greatest Common Factor

The first method for factoring polynomials will be factoring out the greatest common factor. The GCF for a polynomial is the largest monomial that divides each term of the polynomial.

This is like using the distributive law in reverse. The distributive law states that, \(a({b + c}) = ab + ac\). In factoring out the greatest common factor we do this in reverse. \(ab + ac = a({b + c})\). Simply examine each term to see if there is a characteristic common factor in all the terms. If so, we’ll factor the polynomial to get rid of it.

**Step 1: **Identify the GCF of the polynomial.

**Step 2:** Divide the GCF out of every term of the polynomial.

**Example: **\(8{x^4} – 4{x^3} + 10{x^2} = 2{x^2}( {4{x^2} – 2x + 5} )\)

If a term of the polynomial is exactly the same as the GCF when you divide it by the GCF you are left with 1, Not 0. Don’t think, you don’t have anything, there is actually a 1.

You can see it in the following example.

\({x^3}{y^2} + 3{x^4}y + 5{x^5}{y^3} = {x^3}y( {y + 3x + 5{x^2}{y^2}} )\)

### Factoring By Grouping

This method is useful when the polynomial has a lot of terms and there is not one common factor. In some cases, there is not a GCF for ALL the terms in a polynomial. If you have four terms with no GCF, then try factoring by grouping. In this method, the polynomial terms are grouped together to find out their common factors.

**Step 1:** Group the first two terms together and then the last two terms together.

**Step 2: **Factor out a GCF from each separate binomial.

**Step 3: **Factor out the common binomial.

**Example:**

\(3{x^2} – 2x + 12x – 8 = x( {3x – 2} ) + 4( {3x – 2} )= ( {3x – 2} )( {x + 4} )\)

\(4{x^2} + 10x – 6 = 2( {2x – 1} )( {x + 3} )\)

\({x^5} – 3{x^3} – 2{x^2} + 6 = {x^3}( {{x^2} – 3} ) – 2( {{x^2} – 3} ) = ( {{x^2} – 3} )( {{x^3} – 2} )\)

### Factoring using Identities

There are some identities and using them the factorization process becomes much easier. A number of expressions to be factorized are of the form or can be put into the form of identities.

**Standard Algebraic Identities Used for Factoring**

Following standard algebraic identities are factoring formula:

**Identity 1:**\(a^2 – b^2 = ( a + b)(a – b)\)**Identity 2:**\(a^2 + b^2 = (a + b)^2 – 2ab\)**Identity 3:**\(a^3 + b^3 = (a + b)(a^2 + b^2 – ab)\)**Identity 4:**\(a^3 – b^3 = ( a – b)(a^2 + b^2 + ab)\)**Identity 5:**\(a^4 + b^4 = a^4 + 2a^2b^2 + b^4\)**Identity 6:**\(a^4 – b^4 = (a^2 + b^2)(a^2 – b^2)\)

### Factoring by Splitting Terms

A method of factoring a polynomial as a product of two factors by splitting the middle term of a quadratic form basis expression is called the factorization (or factorisation) of a mathematical expression by splitting the middle term.

In order to factorize \(x^2 + bx + c\) we have to find numbers \(p\) and \(q\) such that \(p + q = b\) and \(pq = c\).

After finding \(p\) and \(q\), we split the middle term in the quadratic as \(px + qx\) and get desired factors by grouping the terms.

Example: \(x^2 + 6x + 8\)

In order to factorize \(x^2 + 6x + 8\), we find two numbers \(p\) and \(q\) such that \(p + q = 6\) and \(pq = 8\).

Clearly, 2 + 4 = 6 and 2 × 4 = 8.

We know split the middle term 6x in the given quadratic as \(2x + 4x\), so that

\(x^2 + 6x + 8 = x^2 + 2x + 4x + 8\)

\(= (x^2 + 2x) + (4x + 8)\)

\(= x (x + 2) + 4 (x+ 2)\)

\(= (x + 2) (x + 4)\)

### Long Division of Polynomials

One of the simplest method to find the factors of a polynomial is by dividing it by another known factor just like we do for numbers. This is called as Long Division of Polynomial. Study it in detail in the linked article.

## Factoring Polynomial Theorems

There are some important theorems and properties of polynomials that helps to factor them. It is necessary that we understand them first before moving forwards.

**Bezout’s Theorem**

A polynomial \(P(x)\) is divisible by a binomial \((x – a)\) if and only if \(P(a) = 0\) or \(a\) is the root or zero or solution of a polynomial \(P(x)\) if and only if the polynomial \(P(x)\) is completely divided by \((x – a)\) without any remainders. In other terms, we can say that in general, the number of zeros equals the number of factors of the polynomials. In addition, the number of zeros of a polynomial are always equal to the degree of the polynomial. It is named after Étienne Bézout.

**Remainder Theorem **

Let us understand the Remainder Theorem below:

If \(P(x)\) is divided by \((x – a)\) with a remainder \(r\), then \(P(a) = r\). Let’s see how it works in detail.

When we divide \(f(x)\) by the simple polynomial \(x−a\) we get:

\(P(x) = (x−a)·Q(x) + R(x)\)

\(x−c\) is degree 1, so \(R(x)\) must have degree 0, so it is just some constant \(r\):

\(P(x) = (x−a)·Q(x) + r\)

Now see what happens when we have \(x\) equal to :

\(P(a) = (a − a)·Q(a) + r\)

\(P(a) = (0)·q(c) + r\)

\(P(a) = r\)

**Example: **The remainder after \(2x^2−5x−1\) is divided by \(x−3\)

We don’t need to divide by \((x−3)\) we can just calculate \(f(3)\):

\(2(3)^2−5(3)−1\)

\(= 2\times9−5\times3−1\)

\(= 18−15−1\)

\(= 2\)

And that is the remainder we got from our calculations above. We didn’t need to do Long Division at all!

**Factor Theorem**

The Factor Theorem is a result of the Remainder Theorem, and is based on the same reasoning. A polynomial \(P(x)\) divided by \(Q(x)\) will result in \(R(x)\) with zero remainders if and only if \(Q(x)\) is a factor of \(P(x)\).

If \(x − a\) is indeed a factor of \(P(x)\), then the remainder after division by \(x − a\) will be zero. That is: \(P(x) = (x − a)Q(x)\)

The point of the Factor Theorem is the reverse of the Remainder Theorem. According to the remainder theorem, if you synthetic-divide a polynomial by \(x = a\) and get a zero remainder, then \(x = a\) zero of the polynomial. According to the factor theorem, if you synthetic-divide a polynomial by \(x = a\) and get a zero remainder then, \(x − a\) is a factor of the polynomial

## Solved Examples on Factoring Polynomials

Now that we have learned various things about factoring polynomials. Let’s get some practice on the concepts through some examples.

**Solved Example 1: **Add the polynomials \(5x^2 + 2x – 3\) and \(x^2 + 4x + 4\).

**Solution:**

\((5x^2 + 2x – 3) + (x^2 + 4x + 4)\)

=\(5x^2 + 2x – 3 + x^2 + 4x + 4\)

Bringing the like terms together and writing them in standard form.

=\(5x^2 + x^2 + 2x + 4x – 3 + 4\)

Operating the like terms to get the final answer.

=\(6x^2 + 6x – 1\)

**Solved Example 2: **Find the solution of the polynomial: \(x^4 + 4x^2 =21\)

**Solution:**

Write the polynomial in standard form.

\(x^4 + 4x^2 =21\)

Equate it to zero

\(x^4 + 4x^2 – 21 = 0\)

It is difficult to solve for a fourth power root hence, we substitute \(x^2 = a\). This will make the polynomial quadratic easier to solve.

\(a^2 + 4a – 21 = 0\)

Writing the middle terms such that it becomes easier to factorise the polynomial. See Factorising Quadratic Equations for details.

\(a^2 + 7a – 3a – 21 = 0\)

Grouping the polynomial into two groups of binomials: The first two terms and the last two terms.

\((a^2 + 7a) + (-3a – 21) = 0\)

Find the greatest common factor of each group and write them as their product i.e. reverse of the distributive law.

\(a(a + 7) – 3(a + 7) = 0\)

The two groups in turn have a common factor, we continue factorising.

\((a+7)(a – 3) = 0\)

This is the simplest factor form or prime factorisation of the polynomial. Now we substitute \(x^2 = a\).

Thus, \((x^2+7)(x^2 – 3) = 0\)

\((x^2+7)(x – 3)(x + 3) = 0\) …….. (as \(a^2 – b^2 = (a + b)(a – b)\))

For this equation to exist, either of these factors must be equal to zero.

Hence, \((x^2+7) = 0\) OR \((x – 3) = 0\) OR \((x + 3) = 0\)

\(x^2=-7\), thus \(x=\pm\sqrt{7}i\)

\((x – 3) = 0\), thus \(x = 3 \)

\((x + 3) = 0\), thus \(x = -3\)

Thus, the polynomial will have four solutions at \(x= -\sqrt{7}i, +\sqrt{7}i, -3 and +3\). The number of zeros is equal to the degree of the polynomial.

**Solved Example 3: **Divide the polynomial \(8x^3-10x^2-x+3\) by \((x-1)\).

**Solution:**

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## Factoring Polynomials FAQs

**Q.1 How to factorize quadratic polynomials?**

**Ans.1**In order to factorise a quadratic algebraic expression in the form \(ax^2 + bx + c\) into double brackets: Multiply the end numbers together ( a and c ) then write out the factor pairs of this new number in order. We need a pair of factors that + to give the middle number ( b ) and ✕ to give this new number.

**Q.2 How to factor second degree polynomials?**

**Ans.2**First factor out the greatest common factor (GCF). Try to see if any of the standard identities work. If these two doesn’t work try grouping the terms and find the common factors.

**Q.3 Why do we use factor polynomials?**

**Ans.3**The degree of a linear polynomial is one. Factoring polynomials is the basic step in finding out the solution of the polynomial. The solution of the polynomial or zeros of a polynomial is the values of the variables of the polynomial which when substituted gives us the answer as zero.

**Q.4 How to factor polynomials with 3 terms?**

**Ans.4**For a trinomial, check to see whether it is either of the following forms: \(x^2 + bx + c\): If so, find two integers whose product is c and whose sum is b. \(ax^2 + bx + c\): If so, find two binomials so that The product of the first terms = \(ax^2\); The product of the last terms = \(c\); The sum of outer and inner products = \(bx\)

**Q.5 How to factor polynomials with 4 terms?**

**Ans.5**when a polynomial has four or more terms, the easiest way to factor it is to use grouping.