Factoring Polynomials

Factoring Polynomials refers to factoring a polynomial into irreducible polynomials over a given field. It gives out the factors that together form a polynomial function. A polynomial function is of the form xⁿ + x^{n -1} + x^{n - 2} + . . . . + k = 0, where k is a constant and n is a power. Polynomials are expressions that are formed by adding or subtracting several variables called monomials. Monomials are variables that are formed with a constant and a variable of some degree. Examples of monomials are 5x³, 6a². Monomials having different exponents such as 5x³ and 3x⁴ cannot be added or subtracted but can be multiplied or divided by them. Any polynomial of the form F(a) can also be written as

F(a) = Q(a) x D (a) + R (a)

using Dividend = Quotient x Divisor + Remainder.

If the polynomial F(a) is divisible by Q(a), then the remainder is zero. Thus, F(a) = Q(a) x D(a). That is, the polynomial F(a) is a product of two other polynomials Q(a) and D(a). For example, 2t + 6t² = 2t x (1 + 3t).

Variables, Exponents, Parenthesis and Operations (+, -, x, /) play an important role in factoring a polynomial.

Factorization by dividing the expression by the GCD of the terms of the given expression:

GCD of a polynomial is the largest monomial, which is a factor of each term of the polynomial. It involves finding the Greatest Common Divisor (GCD) or Highest Common Factor (HCF) of the terms of the expression and then dividing each term by its GCD. Therefore the factors of the given expression are the GCD and the quotient thus obtained.

Example 1:

Factorize : 2x³ – 6x² + 4x.

Solution:

Factors of 2x³ are 1, 2, x, x², x³,2x, 2x², 2x³

Factors of 6x² are 1, 2,3, 6, x, x², 2x, 2x² ,3x, 3x² ,6x, 6x²

Factors of 4x are 1,2,4,x,2x,4x.

Thus the GCD of the above terms is 2x.

Dividing 2x³ , -6x² and 4x by 2x, we get x² - 3x + 2

Then the GCD becomes one factor and the quotient is the other factor.

2x³ – 6x² + 4x = 2x .(x² - 3x + 2)

Therefore the factors of 2x³ – 6x² + 4x are 2x and (x² - 3x + 2)

Thus, 2x³ – 6x² + 4x = 2x .(x² - 3x + 2)

Factorization by grouping the terms of the expression:

Grouping the terms of the expression in such a way that there are common factors among the terms of the groups so formed.

Example: Factorize 3x + xy + 3y + y²

Hint: Notice that there is no factor common to all the terms. So regroup the terms of the expression. In this expression, there is a common factor for the first two terms. Similarly the last two terms have a common factor.

Therefore 3x + xy + 3y + y²= x (3 + y) + y ( 3+y)

3x + xy + 3y + y² = ( 3 + y) ( x + y)

This can also be regrouped in another way,

3x + xy + 3y + y² = 3x + 3y + xy + y²

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

3x + xy + 3y + y² = (3 + y) ( x+ y)

Factorizing Trinomial:

Trinomials are the most common polynomials having three terms or three monomials. In this factorization method we do the process of the reverse multiplication of factors (multiplying of binomials is generally done by the FOIL method, which stands for First Outside Inside Last)

To multiply binomials like ( 3x – 5) ( x + 2) by FOIL method we get four terms

( 3x – 5) ( x + 2) = 3x² +6x – 5x -10 = 3x² + x -10.

Thus if a trinomial is given in this form 3x² + x -10 we split the middle term x as 6x - 5x.

3x² + x -10 = 3x² + 6x - 5x -10.

Then we use the method 2 of factoring by grouping.

3x² + 6x - 5x -10 = 3x (x + 2) -5 (x + 2)

3x² + x -10 = (3x - 5 ) ( x + 2)

Quadratic polynomial can be factored using the following three methods:

1. Factoring by grouping

2. completing the square

3. Using quadratic formula.

Some of the identities used while factoring polynomials are

• a² + 2ab + b² = (a + b)²
• a² - 2ab + b² = (a - b)²
• a² - b² = (a + b)(a - b)
• x² + x(a + b) + ab=(x + a)(x + b)
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)

Quadratic equation can be solved using quadratic formula,by substituting the coefficient values in the formula we can find the value of x, If the quadratic equation is ax²+bx+c then x = `(-b+-root(2)(b^(2)-4ac))/(2a)`

Example:

Solve x²-7x+12

Solution:

Given x²-7x+12

a=1,b=-7,c=12

We have x = `(-b+-root(2)(b^(2)-4ac))/(2a)`

x = `(7+-root(2)((-7)^(2)-4*1*12))/(2*1)`

x = `(7+-root(2)(49-48))/(2)`

x = `(7+-1)/2`

x = 8/2 = 4 or x = 6/2 = 3

Solutions is x = 4 and x = 3

This method is used to solve quadratic equation. In this method, the given equation is translated to the nearest square and is solved for the square and will find out the solution. Let us see one example

Example:

Solve using completing the square x²+4x=12

Solution:

x²+4x = 12

Adding 4 on both sides,

x²+4x+4 = 12+4

x²+4x+2² = 16

(x+2)² = 16

x+2 = `sqrt(16)`

x+2 = 4

x = 4-2

x = 2