## How to factor a polynomial

According to our **IB Maths Tutors,** first of all, we need to understand the meaning of factorization. Factorization means writing a higher degree polynomial as a product of linear polynomials.

Suppose we have a quadratic polynomial and we want to factorize it then we have to try to write it as a product of two linear polynomials.

If we have a cubic polynomial and we want to factorize it then we have to try to write it as a product of three linear polynomials. This process continues for all higher degree polynomials.

### Types of polynomial

There are many different types of polynomials classified on the basis of their degree and their number of terms, we have a different way of factorization for almost every type of polynomials

Factorization of a Monomial- Monomial is already a linear polynomial with degree one so we don’t need to factorize it.

### how to factor a polynomial with two terms

A polynomial with two terms is called a binomial. we can have binomials of many types

#### Binomial of degree two

**When both terms have the same signs-** these types of polynomials can’t be factorized, only a few can be factorized using perfect square identities.

W**hen both terms have opposite signs and the power of the variable is divisible by two-**

these polynomials can easily be factorized by using a²-b²=(a+b)(a-b) identity

**Example: 9x²-16y²**

=(3x)²-(4y)²

= (3x-4y)(3x+4y)

we can also factorize polynomials for** degree 4, degree 6, and degree 8** and much more in the same way

**When both terms have opposite signs and the power of the variable is divisible by three- these** polynomials can easily be factorized by using **a³-b³=(a-b)**(a²+ab+b²) or a³+b³=(a+b)(a²-ab+b²) identity

**Example: 64x³-27y³**

** ** =(4x)³-(3y)³

= (3x-4y)(9x²+12xy+16y²)

we can also factorize polynomials for **degree 6, and degree 9** and much more in the same way

### How to factor a polynomial with three terms

A polynomial with three terms is called a cubic polynomial. A trinomial is usually a quadratic trinomial. This can be of two types:

#### A perfect square quadratic trinomial can be solved using this identity

(a+b)²=a²+2ab+b² or by (a-b)²=a²-2ab+b²

**Example- 9x²-24x+16**

=(3x)²-2(3x)(4)+(4)²

=(3x-4)²

#### A generic (non-perfect square) quadratic trinomial then we factorize it using the middle term splitting method.

**Example: 9x²-25x+16**

=9x²-(16x+9x)+16

=9x²-16x-9x+16

=x(9x-16)-1(9x-16)

=(9x-1)96x-1)

**Factorization of cubic polynomials with four terms-these polynomials can be factorized in different ways.**

#### Factorization by using hit and trial method-

we use this method for cubic polynomials of 3 or 4 terms when we have only one variable in the polynomials. hit and trial is used when terms are usually in order

*Example:*

**Find the zeros of f( x) = 2x^{3} + 3x^{2} – 11x – 6**

**Solution:**

**Solution:**

We will find one solution to this polynomial by hit and trial method

Step 1: Use the factor to test the possible values by hit and trial.

f(1) = 2 + 3 – 11 – 6 ≠ 0

f(–1) = –2 + 3 + 11 – 6 ≠ 0

f(2) = 16 + 12 – 22 – 6 = 0

We find that the integer root is 2.

Step 2: Find the other roots either by inspection or by synthetic division. I am showing the inspection method here, you should try division method yourself

2x^{3} + 3*x*^{2} – 11*x* – 6

= (*x* – 2)(*ax*^{2} + *bx + c*)

= (*x* – 2)(2*x*^{2} + *bx + *3)

= (*x* – 2)(2*x*^{2} + 7*x + *3)

= (*x* – 2)(2*x* + 1)(*x* +3)

we have calculated a b and c by inspection or comparison method

**We can use binomial whole cube identity to factorize cubic polynomials that are perfect cubes in itself.**

**(a+b)³=a³+3a²b+3ab²+b³**

We can use this method to factorize cubic polynomials with four terms also but generally, we use it for 2 variables when two terms are perfect cubes and rest two are divisible by 3

**Example: ****27x³+108x²y+144xy³+64y³**

=(3x)³+3(3x)²(4y)+3(3x)(4y)+(4y)³

=(3x+4y)³

#### Besides these methods we can use :

We can also use this method to factorize cubic polynomials with 4 terms. generally, we use it for 2 or 3 variables when 3 terms are perfect cubes and 4th term is divisible by 3

**a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab -bc -ca)**

**If we are ever asked to evaluate or factorize a³+b³+c³ we should first find the sum of a+b+c usually this sum is zero then we can use**

##### a³+b³+c³=3abcd

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