# Ssamacheer Kalvi Math Solution for Exercise 3.11 Part1

This page samacheer kalvi math solution for exercise 3.11 part1 is going to provide you solution for every problems that you find in the exercise no 3.11

## Samacheer Kalvi Math Solution for Exercise 3.11 part1

(1) Simplify the following as a quotient of two polynomials in the simplest form.

(i) [x³/(x - 2)] + [8/(2-x)]

Solution:

= [x³/(x - 2)] + [8/(2-x)]

= [x³/(x - 2)] - [8/(x-2)]

= [ (
x³  - 8)/(x - 2) ]

comparing x³  - 8 with the algebraic identity a³ - b³ = (a - b) (a² + ab+ b²),we get

= (x - 2) (x² + x(2) + 2²)/(x - 2)

= (x² + 2 x + 4)

(ii) (x + 2)/(x² + 3 x + 2)] + (x - 3)/(x² - 2 x - 3)

Solution:

= [(x + 2)/(x² + 3 x + 2)] + [(x - 3)/(x² - 2 x - 3)]

(x² + 3 x + 2) = (x + 1) (x +  2)

(x² - 2 x - 3) = (x - 3) (x + 1) (iii) [(x² - x - 6)/(x² - 9)] + [(x² + 2 x - 24)/(x² - x - 12)]

Solution:

= [(x² - x - 6)/(x² - 9)] + [(x² + 2 x - 24)/(x² - x - 12)]

(x² - x - 6) = (x - 3) (x + 2)

by comparing (x² - 9) with the algebraic identity (a² - b²) = (a + b) (a - b),we get

(x² - 3²) = (x + 3)(x - 3)

(x² + 2 x - 24) = (x + 6) (x - 4)

(x² - x - 12) = (x - 4) (x + 3) In the page samacheer kalvi math solution for exercise 3.11 part1 we are going to see the solution of next problem

(iv) [(x - 2)/(x² - 7 x + 10)] + [(x + 3)/(x² - 2 x - 15)]

Solution:

= [(x - 2)/(x² - 7 x + 10)] + [(x + 3)/(x² - 2 x - 15)]

(x² - 7 x + 10) = (x - 2 ) (x - 5)

(x² - 2 x - 15) = (x - 5) (x + 3) HTML Comment Box is loading comments...