Question 3 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

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Question 3 (i) Factor the following algebraic expressions: (i) 4𝑦^2+1+1/(16𝑦^2 ) Here, we can write 4𝑦^2=(𝟐𝒚)^𝟐 1/(16𝑦^2 )=(𝟏/𝟒𝒚)^𝟐 And, since there +1 i.e. positive sign, we use (a + b)2 Now, 4𝑦^2+1+1/(16𝑦^2 ) = 4𝑦^2+1/(16𝑦^2 )+1 = (𝟐𝒚)^𝟐+(𝟏/𝟒𝒚)^𝟐+𝟐 × 𝟐𝒚 ×𝟏/𝟒𝒚 = (𝟐𝒚+𝟏/𝟒𝒚)^𝟐 Using (𝑎+𝑏)^2 = 𝑎^2 + 𝑏^2 + 2ab Where 𝑎 = 2𝑦, b = 1/4𝑦 Question 3 (ii) Factor the following algebraic expressions: (ii) 9𝑚^2−1/(25𝑛^2 ) We can use 𝒂^𝟐−𝒃^𝟐 formula here 9𝑚^2−1/(25𝑛^2 )=(𝟑𝒎)^𝟐−(𝟏/𝟓𝒏)^𝟐 =(𝟑𝒎+𝟏/𝟓𝒏)(𝟑𝒎−𝟏/𝟓𝒏) Using 𝑎^2−𝑏^2=(𝑎+𝑏)(𝑎−𝑏) Where 𝑎 = 3𝑚, b = 1/5𝑛 Question 3 (iii) Factor the following algebraic expressions: (iii) 27𝑏^3−1/(64𝑏^3 ) We can use 𝒂^𝟑−𝒃^𝟑 formula here 27𝑏^3−1/(64𝑏^3 ) =(𝟑𝒃)^𝟑−(𝟏/𝟒𝒃)^𝟐 =(3𝑏−1/4𝑏)((3𝑏)^2+3𝑏 ×1/4𝑏+(1/4𝑏)^2 ) =(𝟑𝒃−𝟏/𝟒𝒃)(𝟗𝒃^𝟐+𝟑/𝟒+𝟏/(𝟏𝟔𝒃^𝟐 )) Using 𝑎^3−𝑏^3=(𝑎−𝑏)(𝑎^2+𝑎𝑏+𝑏^2) Where 𝑎 = 3𝑏, b = 1/4𝑏 Question 3 (iv) – Method 1 Factor the following algebraic expressions: (iv) 𝑥^2+5𝑥/6+1/6 𝑥^2+5𝑥/6+1/6 Factorising by splitting the middle term = 𝑥^2+𝟏/𝟐 𝒙+𝟏/𝟑 𝒙+1/6 = 𝑥(𝑥+1/2)+1/3 (𝑥+1/2) = (𝒙+𝟏/𝟑)(𝒙+𝟏/𝟐) Splitting the middle term method We need to find two numbers whose Sum = 5/6 Product = 1 × 1/6 = 1/6 Since both sum and product are positive. Thus, both numbers are positive Question 3 (iv) – Method 2 Factor the following algebraic expressions: (iv) 𝑥^2+5𝑥/6+1/6 Question 3 (iv) – Method 2 Factor the following algebraic expressions: (iv) 𝑥^2+5𝑥/6+1/6 𝑥^2+5𝑥/6+1/6 Since there are fraction terms, we multiply and divide by 6 in all terms = 6/6 (𝑥^2+5𝑥/6+1/6) = 1/6 (6𝑥^2+6 × 5𝑥/6+6 × 1/6) = 𝟏/𝟔 (𝟔𝒙^𝟐+𝟓𝒙+𝟏) Factorising by splitting the middle term = 1/6(6𝑥^2+𝟐𝒙+𝟑𝒙+1) Splitting the middle term method We need to find two numbers whose Sum = 5 Product = 6 × 1/6 = 6 Since both sum and product are positive. Thus, both numbers are positive = 1/6 (6𝑥^2+𝟐𝒙+𝟑𝒙+1) = 1/6 (2𝑥(3𝑥+1)+1(3𝑥+1)) = 𝟏/𝟔 (𝟐𝒙+𝟏)(𝟑𝒙+𝟏) = 1/6 × 2(2𝑥/2+1/2) × 3(3𝑥/3+1/3) = 1/6 × 2(𝑥+1/2) × 3(𝑥+1/3) = 6/6 ×(𝑥+1/2)(𝑥+1/3) = (𝒙+𝟏/𝟐)(𝒙+𝟏/𝟑) Question 3 (v) Factor the following algebraic expressions: (v) 27𝑢^3−1/125−(27𝑢^2)/5+9𝑢/25 Here, There are 2 cube terms: 〖27𝑛〗^3=(3𝑛)^3 and 1/125=(1/5)^3 Since it is (−𝟏)/𝟏𝟐𝟓 is negative, thus we use (a – b)3 Now, 27𝑢^3−1/125−(27𝑢^2)/5+9𝑢/25 = 27𝑢^3−(27𝑢^2)/5+9𝑢/25−1/125 =(𝟑𝒖)^𝟑−𝟑 × (𝟑𝒖)^𝟐 ×𝟏/𝟓+𝟑 × 𝟑𝐮 × (𝟏/𝟓)^𝟐−(𝟏/𝟓)^𝟑 =(𝟑𝐮−𝟏/𝟓)^𝟑 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 3𝑢, 𝑏 = 1/5 Question 3 (vi) Factor the following algebraic expressions: (vi) 64𝑦^3+1/125 𝑧^3 We can use 𝒂^𝟑+𝒃^𝟑 formula here 64𝑦^3+1/125 𝑧^3 =(𝟒𝒚)^𝟑+(𝟏/𝟓 𝒛)^𝟐 =(4𝑦+1/5 𝑧)((4𝑦)^2−4𝑦 ×1/5 𝑧+(1/5 𝑧)^2 ) =(𝟒𝒚+𝟏/𝟓 𝒛)(𝟏𝟔𝒚^𝟐−𝟒/𝟓 𝒚𝒛+𝟏/𝟐𝟓 𝒛^𝟐 ) Using 𝑎^3+𝑏^3=(𝑎+𝑏)(𝑎^2−𝑎𝑏+𝑏^2) Where 𝑎 = 4𝑦, b = 1/5 𝑧 Question 3 (vii) Factor the following algebraic expressions: (vii) 𝑝^3+27𝑞^3+𝑟^3−9𝑝𝑞𝑟 Since there are 3 cubes, we use 𝒙^𝟑+𝒚^𝟑+𝒛^𝟑formula Now, 𝑝^3+27𝑞^3+𝑟^3−9𝑝𝑞𝑟 = 𝒑^𝟑+(𝟑𝒒)^𝟑+𝒓^𝟑−𝟑 × 𝒑 × 𝟑𝒒 × 𝒓 = (𝑝+3𝑞+𝑟)(𝑝^2+(3𝑞)^2+𝑟^2− 𝑝 × 3𝑞 −𝑝 × 𝑟−3𝑞 × 𝑟) = (𝒑+𝟑𝒒+𝒓)(𝒑^𝟐+𝟗𝒒^𝟐+𝒓^𝟐−𝟑𝒑𝒒 −𝒑𝒓−𝟑𝒒𝒓) Using 𝒙^𝟑+𝒚^𝟑+𝒛^𝟑−𝟑𝒙𝒚𝒛=(𝒙+𝒚+𝒛)(𝒙^𝟐+𝒚^𝟐+𝒛^𝟐−𝒙𝒚−𝒚𝒛−𝒛𝒙) Putting x=𝑝,𝑦=3𝑞,𝑧=𝑟 Question 3 (viii) Factor the following algebraic expressions: (viii) 9𝑚^2−12𝑚+4 Here, we can write 9𝑚^2=(𝟑𝒎)^𝟐 4=𝟐^𝟐 And, since there −12𝑚 i.e. negative sign, we use (a – b)2 Now, 9𝑚^2−12𝑚+4 = 9𝑚^2+4−12𝑚 = (𝟑𝒎)^𝟐+𝟐^𝟐−𝟐 × 𝟑𝒎 × 𝟐 Using (𝑎−𝑏)^2 = 𝑎^2 + 𝑏^2 – 2ab Where 𝑎 = 3𝑚, b = 2 =(𝟑𝒎−𝟐)^𝟐 Question 3 (ix) Factor the following algebraic expressions: (ix) 9𝑥^3−8/3 𝑦^3+𝑧^3/3+6𝑥𝑦𝑧 Since there are 3 cubes, we use 𝒙^𝟑+𝒚^𝟑+𝒛^𝟑formula But, 𝟗𝒙^𝟑 isn’t a cube for any number, but 𝟐𝟕𝒙^𝟑 is So, we multiply and divide the full expression with 3 Now, 9𝑥^3−8/3 𝑦^3+𝑧^3/3+6𝑥𝑦𝑧 = 𝟑/𝟑 (𝟗𝒙^𝟑−𝟖/𝟑 𝒚^𝟑+𝒛^𝟑/𝟑+𝟔𝒙𝒚𝒛) = 1/3 (3 × 9𝑥^3−3 × 8/3 𝑦^3+3 × 𝑧^3/3+3 × 6𝑥𝑦𝑧) = 𝟏/𝟑 (𝟐𝟕𝒙^𝟑−𝟖𝒚^𝟑+𝒛^𝟑+𝟏𝟖𝒙𝒚𝒛) Factorising the expression now = 1/3 ((3𝑥)^3+(−2𝑦)^3+𝑧^3−3 × 3𝑥 × (−2𝑦) × 𝑧 ) = 1/3 (3𝑥+(−2𝑦) +𝑧) (〖(3𝑥)〗^2+(−2𝑦)^2+𝑧^2−3𝑥 × (−2𝑦) −3𝑥 × 𝑧−(−2𝑦) × 𝑧) = 1/3 (𝟑𝒙−𝟐𝒚+𝒛)(𝟗𝒙^𝟐+𝟒𝒚^𝟐+𝒛^𝟐+𝟔𝒙𝒚−𝟑𝒙𝒛+𝟐𝒚𝒛) Using 𝒙^𝟑+𝒚^𝟑+𝒛^𝟑−𝟑𝒙𝒚𝒛=(𝒙+𝒚+𝒛)(𝒙^𝟐+𝒚^𝟐+𝒛^𝟐−𝒙𝒚−𝒚𝒛−𝒛𝒙) Putting x=3𝑥,𝑦=−2𝑦,𝑧=𝑧 Question 3 (x) Factor the following algebraic expressions: (x) 4𝑥^2+9𝑦^2+36𝑧^2+12𝑥𝑧+36𝑦𝑧+24𝑥𝑦 Here, There are 3 square terms, so we use (a + b + c)2 Our square terms would be 𝟒𝒙^𝟐, 𝟗𝒚^𝟐 , 〖𝟑𝟔𝒛〗^𝟐 Now, 4𝑥^2+9𝑦^2+36𝑧^2+12𝑥𝑧+36𝑦𝑧+24𝑥𝑦 = (𝟐𝒙)^𝟐+(𝟑𝒚)^𝟐+〖(𝟔𝒛)〗^𝟐 + 𝟐 × (𝟐𝒙)×(𝟑𝒚) + 𝟐 × (𝟐𝒙) ×(𝟔𝒛) +𝟐 ×(𝟑𝒚) × (𝟔𝒛) Using (𝑎+𝑏+𝑐)^2=𝑎^2+𝑏^2+𝑐^2+2𝑎𝑏+2𝑏𝑐+2𝑎𝑐 Putting 𝑎 = 2𝑥, 𝑏 = 3𝑦 & 𝑐 = 6𝑧 = (𝟐𝒙+𝟑𝒚+𝟔𝒛)^𝟐 Question 3 (xi) Factor the following algebraic expressions: (xi) 27𝑢^3−1/216−(9𝑢^2)/2+𝑢/4 Here, There are 2 cube terms: 〖27𝑛〗^3=(3𝑛)^3 and 1/216=(1/6)^3 Since it is (−𝟏)/𝟐𝟏𝟔 is negative, thus we use (a – b)3 Now, 27𝑢^3−1/216−(9𝑢^2)/2+𝑢/4 = 27𝑢^3−(9𝑢^2)/2+𝑢/4−1/216 =(𝟑𝒖)^𝟑−𝟑 × (𝟑𝒖)^𝟐 ×𝟏/𝟔+𝟑 × 𝟑𝐮 × (𝟏/𝟔)^𝟐−(𝟏/𝟔)^𝟑 =(𝟑𝐮−𝟏/𝟔)^𝟑 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 3𝑢, 𝑏 = 1/6