Factoring a Polynomial Factoring a Polynomial To know the ways of Factoring a Polynomial you should be familiar with the term polynomial. A polynomial is an algebraic equation. In this the variables involved can have only non negative integral powers. Example of polynomial are :f ( x ) = 4x + 1 / 2 g ( y ) = 3y2 – ( 5 / 3 ) y + 9 p ( x ) = 6x3 – 4x2 + x – 1 / √2 q ( u ) = 3u5 – ( 2 / 3 ) u4 + u2 – 1 / 2. We use a factor theorem for Factoring a Polynomial. The theorem is :Let f ( x ) be a polynomial of degree n > 1 and let a be any real number. 1. If f ( a ) = 0 then ( x – a ) is a factor of f ( x ). 2. If ( x – a ) is a factor of f ( x ) then f ( a ) = 0. Proof :- 1. Let f ( a ) = 0

Know More About :- Pythagorean Identities

Tutorcircle.com

Page No. : 1/4

On dividing f ( x ) by ( x - a ), let g ( x ) be the quotient. Also, by the remainder theorem, when f ( x ) is divided by ( x - a ), then the remainder is f ( a ). f ( x ) = ( x – a) . g ( x ) + f ( a ) => f ( x ) = ( x – a) . g ( x ) [ where f ( a ) = 0 ( given ) ] => x – a is factor of f ( x ). Proof :- 2. Let ( x – a ) be a factor of f ( x ). On dividing the f ( x ) by ( x - a ), let g ( x ) be the quotient. Then, f ( x ) = ( x – a ) . g ( x ) => f ( a ) = 0 [ putting x = a ] Thus, ( x – a ) is a factor of f ( x ) => f ( a ) = 0 A polynomial g ( x ) is called a factor of the polynomial p ( x ), if g ( x ) divides p ( x ) exactly. Examples :( i ) ( x – 2 ) is a factor of ( x2 – 3x - 10 ) ( ii ) ( x + 1 ) is a factor of ( x2 + 10x + 9 ) The process to express a given polynomial as the product of polynomial, each of degree less than that of the given polynomial such that no such a factor has a factor of lower degree is called factorization. Example :-

Learn More :- Reference Angle

Tutorcircle.com

Page No. : 2/4

( i ) ( x2 – 16 ) = ( x - 4 ) ( x + 4 ) ( ii ) ( x2 – 3x + 2 ) = ( x – 2 ) ( x – 1 ) We will find its factor by Quadratic Formula. Formula to find factors is given by: P = -b + √ (b2 - 4ac) / 2a, here value of 'a' is 1, value of 'b' is 1 and value of 'c' is -4. So put these values in formula. On putting these values we get: P = - 1 + √ [(1)2 - 4(1) (-4)] / 2(1); on moving ahead we get: P = - 1 + √ (1 + 16) / 2, we can also write it as: P = - 1 + √ (17) / 2. So here we get two factor of this expression, one positive and other negative. P = -1 + √ (17) / 2 and P = -1 - √ (17) / 2. These two are factors of above expression. Using this formula we can find factors of any polynomial expression.

Tutorcircle.com

Page No. : 2/3 Page No. : 3/4

Thank You For Watching

Presentation