Question 2 - Figure it out - Page 154-156 - Chapter 6 Class 8 - We Distribute yet things Multiply (Ganita Prakash)

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Question 2 Use either a suitable identity or the distributive property to find each of the following products. (i) (p – 1) (p + 11)We will use distributive property for this (𝒑−𝟏) (𝒑+𝟏𝟏)=𝑝(𝑝+11)−1(𝑝+11) =𝑝^2+11𝑝−𝑝−11 =𝒑^𝟐+𝟏𝟎𝒑−𝟏𝟏 Question 2 Use either a suitable identity or the distributive property to find each of the following products. (ii) (3a – 9b) (3a + 9b)We can use the property (x + y)(x – y) = x2 – y2 Now, (3a – 9b) (3a + 9b) = (3a)2 – (9b)2 = 32 × a2 – 92 × b2 = 9a2 – 81b2 Question 2 Use either a suitable identity or the distributive property to find each of the following products. (iii) –(2y + 5) (3y + 4)We will use distributive property for this First, we calculate bracket and then multiply negative sign Now, −(𝟐𝐲+𝟓)(𝟑𝐲+𝟒)=−[2𝑦(3𝑦+4)+5(3𝑦+4)] =−[2𝑦(3𝑦+4)+5(3𝑦+4)] =−[2𝑦 × 3𝑦+8𝑦+15𝑦+20] =−[6𝑦^2+23𝑦+20] =−𝟔𝒚^𝟐−𝟐𝟑𝒚−𝟐𝟎 Question 2 Use either a suitable identity or the distributive property to find each of the following products. (iv) (6x + 5y)2We can use the property (a + b)2 = a2 + 2ab + b2 Now, (𝟔𝒙+𝟓𝒚)^𝟐=(6𝑥)^2+2 × 6𝑥 × 5𝑦+(5𝑦)^2 =6^2 × 𝑥^2+2 × 6 × 5 × 𝑥𝑦+5^2 × 𝑦^2 =𝟑𝟔𝒙^𝟐+𝟔𝟎𝒙𝒚+𝟐𝟓𝒚^𝟐 Question 2 Use either a suitable identity or the distributive property to find each of the following products. (v) (2x – 1/2)2We can use the property (a – b)2 = a2 – 2ab + b2 Now, (𝟐𝒙−𝟏/𝟐)^𝟐=(2𝑥)^2−2 × 2𝑥 ×1/2+(1/2)^2 =2^2 × 𝑥^2−2 × 2 ×1/2 × 𝑥+1/4 =𝟒𝒙^𝟐−𝟐𝒙+𝟏/𝟒 Question 2 Use either a suitable identity or the distributive property to find each of the following products. (vi) (7p) × (3r) × (p + 2)First we multiply 7p & 3r, And then we distribute in the bracket Now, 7𝑝 × 3𝑟 × (𝑝+2)=(7 × 3) ×(𝑝 × 𝑟) ×(𝑝+2) =𝟐𝟏𝒑𝒓 (𝒑+𝟐) =21𝑝𝑟 × 𝑝+21𝑝𝑟 ×2 =𝟐𝟏𝒑^𝟐 𝒓+𝟒𝟐𝒑𝒓