TECHNICS OF FACTORIZATION

Polynomial – A polynomial is an algebraic expression which has a power of non negative integer such as →2, x, x², x² +1, 2x³ + 3x +4 etc.

Polynomial degree

2 0

X 1

x² 2

x² +1 2

2x³ + 3x +4 3

Factor of polynomial→ When a polynomial is divisible by another polynomial then the later will be the factor of the polynomial. As an example 2x² +x is divisible by x and 2x +1.There are different technics of evaluating factors of polynomials, depending on its structure.

Examples: Evaluate the factors of the following polynomial.

→First method is to get the common terms between two or more terms.

(i) x²+ xy

= x .x + x.y

Hint : x is common between both the terms

= x( x + y)

So, x and (x + y) will be the factor of x²+ xy.

Factorize the following algebric expression in similar ways.

(a) 4x² + 8xy (b) 8xy + xy² (c) 9x³ + 12x²y³

→Second method is to make pair of two terms and get the common between them.

(ii) x² + xy + x²y + y²x

= (x² + xy) + (x²y + y²x)

=x(x + y) + xy( x + y)

= (x + y) (x + xy)

=x(1 +y)(x+ y)[ applying the first method on (x + xy)]

Factorize the following polynomials in similar way.

(a) xy³ + x³y - x³y² - xy⁴ (b) 3xy² + 4x²y + 27x³y² +36x⁴y

→USING THE IDENTITIES WHICH ARE SHOWN BELLOW:

IMPORTANT ALGEBRIC IDENTITIES

· (x +y)² = x² + y² + 2xy

· (x ─ y)² = x² + y² ─2xy

· ( x + a)(x +b) = x² + (a + b)x + ab

· X² ─ y² = (x + y)(x ─y)

· ( x + y)³ = x³ + y³ + 3x²y + 3y²x

· (x ─ y)³ = x³ ─ y³ ─3x²y + 3y²x

· (x + y + z )² = x² + y² + z² + 2xy + 2yz + 2zx

· X³ + y³ + z³ ─ 3xyz = (x + y + z )( x² + y² + z² -2xy - 2yz -2xz)

· If x + y + z = 0 then X³ + y³ + z³ = 3xyz

· x³ + y³ = (x + y )( x² + y²─ xy)

· x³ ─ y³ = (x ─ y )( x² + y²+ xy)

(x +y)² = x² + y² + 2xy

Factorize The Following polynomial:

16 x² + 24xy + 9y²

Applying the identity (x + y)² = x² + y² + 2xy

=(4x)² + 2.4x .3y + (3y)²

=(4x + 3y)²= (4x + 3y)(4x + 3y)

Factorize The Polynomial:

16 x² ─ 24xy + 9y²

(4x)² ─2.4x.3y + (3y)²

(4x ─3y)²

x² ─ y² = (x + y)(x ─y)

Example: Factorize the following polynomial:

16x² ─ 4y²

(4x)² ─ (3y)²

(4x + 3y)( 4x ─ 3y)

Questions for practice:

(i) 9 ─ 16a² (ii) x⁴ ─ y⁴ (iii) 16a⁴ ─ 81b⁴

( x + y)³ = x³ + y³ + 3x²y + 3y²x

Factorize the following polynomial:

8x³ + 27y³ + 36x²y + 54y²x

Applying the identity ( x + y)³ = x³ + y³ + 3x²y + 3y²x

=(2x)³ + (3y)³ + 36x²y + 54y²x

=( 2x + 3y)³

=( 2x + 3y)( 2x + 3y)( 2x + 3y)

Questions for practice:

(i) 27y³ +27y²z + 9z²y + z³ (ii) 125a³ +300a²b + 240b²a + 64b³

(x ─ y)³ = x³ ─ y³ ─3x²y + 3y²x

Factorize the following:

8x³ ─ 27y³ ─ 36x²y + 54y²x

=(2x)³─(3y)³ ─ 36x²y + 54y²x

=(2x ─ 3y)³( applying the above mentioned identity)

=(2x ─ 3y)(2x ─ 3y)(2x ─ 3y)

Questions for practice :

(i) 27y³ ─27y²z + 9z²y ─z³ (ii) 125a³ ─300a²b + 240b²a ─ 64b³

(x + y + z )² = x² + y² + z² + 2xy + 2yz + 2zx

Factorize the following:

· 16a² + 25b² + 9c² ─40ab ─30bc + 24ac

= (4a)² + (5b)² + (3c)² ─40ab ─30bc + 24ac

The terms ─40ab, ─30bc are negative and b is common between them so the term 5b must be ─5b.

=(4a)² + (─5b)² + (3c)² ─40ab ─30bc + 24ac

=(4a ─5b + 3c)²

= (4a ─5b + 3c)(4a ─5b + 3c)

Questions for practice :

(i) 25x² + 9y² + 16z² + 30xy + 24yz + 40zx

(ii) 4x² + y² + 8z² ─4xy ─4 yz+ 8 xz

X³ + y³ + z³ ─ 3xyz = (x + y + z )( x² + y² + z² ─2xy ─2yz ─2xz)

Factorize the following expression:

· 27x³ + 8y³ + z³ ─18xyz

= (3x)³ + (2y)³ + z³ ─18xyz

Applying the above mentioned identity.

=( 3x + 2y + z) [ (3x)² + (2y)² + z²─12xy─4yz ─ 6zx]

=( 3x + 2y + z) ( 9x² + 4y² + z²─12xy ─ 4yz ─ 6zx)

Questions for practice :

(i) 125a³ + 8b³ + 27c³ ─90abc (ii) 64x³ ─ 27y³ + z³ + 36xyz

(ii) Evaluate the following sum without calculating the cubes .

· 125³ + (─50)³ + (─75)³

SPLIT UP METHODS OF FACTORIZTIONS→We can evaluate the factors of the quadratic polynomial of the expression: Ax² + Bx + C, through this method.In this method one has to choose two factors such that their product is AXC and the sum or difference of them is B.

Examples :

· 5a² + 8a ─ 4

Selecting two factors of the product 5x4 in such a way that their sum or difference is 8 and product is 5x4=5x2x2 = 10x2, it comes 10─2 =8

= 5a² + (10─2)a ─ 4

= 5a² +10a ─2a ─ 4

= 5a(a + 2) ─2(a + 2)

= (a +2)(5a ─ 2)

Questions for practice:

(a) 2y² +7y +3

(b) 6t² + 5t─6

Factorization through remiendr’s theoram→In this method we have to find out one or more factors through the trial basis and then dividing the polynomial by the product of the factors so that a quotient is quadratic ,polynomial, then factorizing the quadratic polynomial by the split up method.

Factorize the following polynomial:

· 5x³ + 13x² + 4x ─ 4

The product of first and last term is 5x4 = 20, one or more than one factor of 20 will be the solution of this polynomial as an ex. 1, ─1, 2,─2..etc

Putting the value of x = ─1

5(─1)³ + 13(─1)² + 4(─1) ─4

= 0

Therefore ( x + 1) will be one of the factor of the given polynomial.

So ,dividing the given polynomial by (x +1).

(5x³ + 13x² + 4x ─ 4) (x +1)

= 5x² + 8x─ 4

= 5x² + 10x ─ 2x ─4

= 5x( x + 2) ─ 2(x + 2)