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Example 2 - Factorise 10x^2 - 18x^3 + 14x^4 - Chapter 14 Class 9

Example 2 - Chapter 14 Class 8 Factorisation - Part 2
Example 2 - Chapter 14 Class 8 Factorisation - Part 3


Transcript

Example 2 (Method 1) Factorise 10π‘₯^2 – 18π‘₯^3 + 14π‘₯^4 10π‘₯^2 = 10 Γ— π‘₯^2 = 2 Γ— 5 Γ— π‘₯^2 = 2 Γ— 5 Γ— π‘₯ Γ— π‘₯ 18π‘₯^3 = 18 Γ— π‘₯^3 = 2 Γ— 3 Γ— 3 Γ— π‘₯^3 = 2 Γ— 3 Γ— 3 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ 14π‘₯^4 = 14 Γ— π‘₯^4 = 2 Γ— 7 Γ— π‘₯^4 = 2 Γ— 7 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ Now, 10π‘₯^2 = 2 Γ— 5 Γ— π‘₯ Γ— π‘₯ 18π‘₯^3 = 2 Γ— 3 Γ— 3 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ 14π‘₯^2 = 2 Γ— 7 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ So, 2, π‘₯, π‘₯ are common factors. 10π‘₯^2 + 18π‘₯^3+14π‘₯^4 = (2 Γ— 5 Γ— π‘₯ Γ— π‘₯) βˆ’ (2 Γ— 3 Γ— 3 Γ— π‘₯ Γ— π‘₯" Γ— " π‘₯) + (2 Γ— 7 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ Γ— π‘₯) = 2 Γ— π‘₯ Γ— π‘₯ (5 βˆ’ (3 Γ— 3 Γ— π‘₯) + (7 Γ— π‘₯ Γ— π‘₯)) = 2π‘₯^2 (5 βˆ’ 9π‘₯ + 7π‘₯^2) = 2𝒙^𝟐 (7𝒙^𝟐 βˆ’ 9𝒙 + 5) Example 2 (Method 2) Factorise 10π‘₯^2 – 18π‘₯^3 + 14π‘₯^410π‘₯^2 – 18π‘₯^3 + 14π‘₯^4 = (2 Γ— 5π‘₯^2) βˆ’ (2 Γ— 9π‘₯^3) + (2 Γ— 7π‘₯^4) Taking 2 common from all the terms = 2 (5π‘₯^2 βˆ’ 9π‘₯^3 + 7π‘₯^4) = 2 ( (5 Γ— π‘₯^2) βˆ’ (9π‘₯ Γ— π‘₯^2) + (7π‘₯^2 Γ— π‘₯^2) ) Taking π‘₯^2 common from all the terms = 2π‘₯^2 (5 βˆ’ 9π‘₯ + 7π‘₯^2) = 2𝒙^𝟐 (7𝒙^𝟐 βˆ’ 9𝒙 + 5)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.