
Last updated at Dec. 24, 2018 by Teachoo
Transcript
Example 10 Simplify (a + b) (2a β 3b + c) β (2a β 3b) c. Here there are two expressions 1st expression = (π+π) (2πβ3π+π) 2nd expression = (2πβ3π) π Solving 1st expression (π+π) (2πβ3π+π) = π(2πβ3π+π)+π(2πβ3π+π) = (πΓ2π)β(πΓ3π)+(πΓπ)+(πΓ2π)β(πΓ3π)+(πΓπ) = 2π^2β3ππ+ππ+2ππβ3π^2+ππ = 2π^2β3π^2+ππ+ππβ3ππ +2ππ = 2π^2β3π^2+ππ+ππβππ Solving 2nd expression (2πβ3π)π = 2ππβ3ππ Now, our expression is (π+π)(2πβ3π+π)β(2πβ3π)π = 2π^2β3π^2+ππ+ππβππβ(2ππβ3ππ) = 2π^2β3π^2+ππ+ππβππβ2ππ+3ππ = 2π^2β3π^2βππ+(ππβ2ππ)+(ππ+3ππ) = ππ^πβππ^πβππβππ+πππ
Multiplication of Polynoimals by Polynomials
Multiplying Binomial by a Trinomial
Ex 9.4, 1 (i)
Ex 9.4, 1 (ii)
Ex 9.4, 1 (iv) Important
Ex 9.4, 1 (iii) Important
Ex 9.4, 2 (i)
Ex 9.4, 2 (ii)
Example 8
Example 9 Important
Ex 9.4, 2 (iii)
Ex 9.4, 2 (iv)
Ex 9.4, 1 (v) Important
Ex 9.4, 1 (vi) Important
Ex 9.4, 3 (iii)
Ex 9.4, 3 (i)
Ex 9.4, 3 (ii)
Ex 9.4, 3 (vi)
Ex 9.4, 3 (viii)
Example 10 Important You are here
Ex 9.4, 3 (iv) Important
Ex 9.4, 3 (v) Important
Ex 9.4, 3 (vii) Important
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