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Ex 12.2, 4 (iv) - Chapter 12 Class 8 Factorisation
Last updated at Aug. 8, 2023 by Teachoo
Ex 12.2
Ex 12.2, 1 (i)
Ex 12.2, 1 (ii)
Ex 12.2, 1 (iii) Important
Ex 12.2, 1 (iv) Important
Ex 12.2, 1 (v)
Ex 12.2, 1 (vi)
Ex 12.2, 1 (vii) Important
Ex 12.2, 1 (viii)
Ex 12.2, 2 (i)
Ex 12.2, 2 (ii) Important
Ex 12.2, 2 (iii)
Ex 12.2, 2 (iv) Important
Ex 12.2, 2 (v)
Ex 12.2, 2 (vi)
Ex 12.2, 2 (vii) Important
Ex 12.2, 2 (viii) Important
Ex 12.2, 3 (i)
Ex 12.2, 3 (ii) Important
Ex 12.2, 3 (iii)
Ex 12.2, 3 (iv) Important
Ex 12.2, 3 (v) Important
Ex 12.2, 3 (vi)
Ex 12.2, 3 (vii)
Ex 12.2, 3 (viii) Important
Ex 12.2, 3 (ix)
Ex 12.2, 4 (i)
Ex 12.2, 4 (ii) Important
Ex 12.2, 4 (iii)
Ex 12.2, 4 (iv) Important You are here
Ex 12.2, 4 (v) Important
Ex 12.2, 5 (i)
Ex 12.2, 5 (ii) Important
Ex 12.2, 5 (iii)
Chapter 12 Class 8 Factorisation
Serial order wise
Transcript
Ex 12.2, 4 Factorise. (iv) 𝑥^4 – 〖"(x – z)" 〗^4𝑥^4 – 〖"(x – z)" 〗^4 = 〖(𝑥^2)〗^2 − [(𝑥 − z)^2 ]^2 Using 𝒂^𝟐 − 𝒃^𝟐 = (a + b) (a − b) Here 𝑎=𝑥^2 and b = (𝑥 − z)^2 = [𝑥^2+(𝑥 − z)^2 ] [𝑥^2 −(𝑥 − z)^2] Using 𝒂^𝟐 − 𝒃^𝟐 = (a + b) (a − b) Here 𝑎=𝑥 and b = (𝑥 − z) = [𝑥^2+(𝑥 − z)^2 ] [𝑥+(𝑥 − z)] [𝑥 − (𝑥 − z)] = [𝑥^2+(𝑥 − z)^2 ] [𝑥+ 𝑥 − z] [𝑥 −𝑥+𝑧] = [𝑥^2+(𝑥 − z)^2 ] [2𝑥− z] [𝑧] = [𝒙^𝟐+(𝒙^𝟐+𝒛^𝟐 − 𝟐𝒙𝒛)][𝟐𝒙− 𝒛] [𝒛] = [𝑥^2+𝑥^2+𝑧^2 − 2𝑥𝑧][2𝑥− z] [𝑧] = [〖2𝑥〗^2+𝑧^2 − 2𝑥𝑧][2𝑥− z] [𝑧] = 𝐳 (𝟐𝒙−𝒛) (𝟐𝒙^𝟐+𝒛^𝟐 − 𝟐𝒙𝒛) Using 𝒂^𝟐 − 𝒃^𝟐 = 𝑎^2+𝑏^2 − 2ab Here 𝑎=𝑥 and b = 𝑧
