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Ex 2.4 Q8 (i,ii,iii) class 9 Maths New NCERT | Class 9 Polynomials Ex 2.4 q8 | EX 2.5 q8 class 9
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Ex 2.4 Q8 (i,ii,iii) class 9 Maths New NCERT | Class 9 Polynomials Ex 2.4 q8 | EX 2.5 q8 class 9
In this video we have solved part 1,2 and 3 of Question 8 Exercise 2.4 from ncert new syllabus chapter 2 Polynomials class 9th. In old ncert book it is given as Question number 8 Exercise 2.5
Identities used:
(a+b)³=a³+b³+3a²b+3ab²
(a-b)³= a³-b³-3a²b+3ab²
Q8. Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³– 12a²b + 6ab²
(iii) 27 – 125a³– 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108ab²
(v) 27p³– 1/216- 9/ 2p²+1/4 p
click here 👇 for
#factorise
#polynomialsclass9
#factorisation
#algebraicidentities
#algebraformula
#divisionofpolynomials
#factortheorem
#polynomials
#9thmaths
#ncertmathsclass9
#polynomialclass9
#polynomialsclass9th
#maths
#mathematicsclass9
In this video we have solved part 1,2 and 3 of Question 8 Exercise 2.4 from ncert new syllabus chapter 2 Polynomials class 9th. In old ncert book it is given as Question number 8 Exercise 2.5
Identities used:
(a+b)³=a³+b³+3a²b+3ab²
(a-b)³= a³-b³-3a²b+3ab²
Q8. Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³– 12a²b + 6ab²
(iii) 27 – 125a³– 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108ab²
(v) 27p³– 1/216- 9/ 2p²+1/4 p
click here 👇 for
#factorise
#polynomialsclass9
#factorisation
#algebraicidentities
#algebraformula
#divisionofpolynomials
#factortheorem
#polynomials
#9thmaths
#ncertmathsclass9
#polynomialclass9
#polynomialsclass9th
#maths
#mathematicsclass9
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