Factorize: x 4 - 8 x 2 + 16

Given, x 4 - 8 x 2 + 16 A quadratic expression a x 2 + b x + c can be factorised by splitting the middle term b such that b is sum of two numbers whose product is equal to a c . Here, a c = 1 × 16 = 16 b = - 8 = - 4 - 4 ∵ - 4 × - 4 = 16 Therefore, x 4 - 8 x 2 + 16 = x 4 - 4 x 2 - 4 x 2 + 16 = x 2 x 2 - 4 - 4 x 2 - 4 = x 2 - 4 x 2 - 4 = x 2 - 2 2 x 2 - 2 2 = x + 2 x - 2 x + 2 x - 2 ∵ a 2 - b 2 = a + b a - b = x + 2 2 x - 2 2 Hence, x 4 - 8 x 2 + 16 = x + 2 2 x - 2 2

Divide: 15 x 2 + 10 x + 21 y + 1 ÷ 5 y + 1 x + 7

Given 15 x 2 + 10 x + 21 y + 1 ÷ 5 y + 1 x + 7 . Here we need to divide algebraic expression of a polynomial with another polynomial. Now we have to factorize the terms given, we get, 15 x 2 + 10 x + 21 y + 1 ÷ 5 y + 1 x + 7 = 15 x 2 + 10 x + 21 y + 1 5 y + 1 x + 7 ⇒ 15 x 2 + 10 x + 21 y + 1 5 y + 1 x + 7 = 15 x 2 + 3 x + 7 x + 21 y + 1 5 y + 1 x + 7 ∵ 21 = 3 × 7 , 10 = 3 + 7 [Splitting middle term inside the bracket] ⇒ 15 x 2 + 10 x + 21 y + 1 5 y + 1 x + 7 = 15 x x + 3 + 7 x + 3 y + 1 5 y + 1 x + 7 [Taking out common factors from each of two terms inside the bracket at numerator] ⇒ 15 x 2 + 10 x + 21 y + 1 5 y + 1 x + 7 = 15 x + 3 x + 7 y + 1 5 y + 1 x + 7 [Taking the binomial common factor to outside the bracket] ⇒ 15 x 2 + 10 x + 21 y + 1 5 y + 1 x + 7 = 3 x + 3 [Cancelling the equal factor terms both in numerator and denominator] Hence, 15 x 2 + 10 x + 21 y + 1 ÷ 5 y + 1 x + 7 = 3 x + 3

Write the missing term: 3 x y 3 x + y y - 3 x ÷ 3 y - 3 x = _ _ _ _ _

Given 3 x y 3 x + y y - 3 x ÷ 3 y - 3 x = _ _ _ _ _ . Here we need to find the missing term in the given algebraic equation. From the question, it is very clear that it is the division of a polynomial with another polynomial. So, 3 x y 3 x + y y - 3 x ÷ 3 y - 3 x = 3 x y 3 x + y y - 3 x 3 y - 3 x ⇒ 3 x y 3 x + y y - 3 x 3 y - 3 x = 3 × x y 3 x + y × y - 3 x 3 × y - 3 x ⇒ 3 x y 3 x + y y - 3 x 3 y - 3 x = x y 3 x + y Thus, we have 3 x y 3 x + y y - 3 x ÷ 3 y - 3 x = x y 3 x + y Hence, the missing term is x y 3 x + y .

Write the missing term: x 2 y + x y 2 x y = _ _ _ _ _

Given x 2 y + x y 2 x y = _ _ _ _ _ . Here we need to find the missing term in the given algebraic equation. From the question, it is very clear that it is the division of a polynomial with another polynomial. So, x 2 y + x y 2 x y = x y x + y x y [Taking the common factors outside from the bracket] ⇒ x y x + y x y = x y x + y x y i.e., x y x + y x y = x + y x 2 y + x y 2 x y = x + y Hence, the missing term is x + y .

Write the missing term: Dividend = _____ × _____ + Remainder

Given Dividend = _____ × _____ + Remainder . Here we need to find the missing part. From the question, it is very clear that it is the related to division of terms. For example, when 6 x 3 is divided by 2 x , we get 3 x 2 as a quotient and 0 as a remainder. i.e., 6 x 3 can be written as, 6 x 3 = 3 x 2 × 2 x or 6 x 3 = 3 x 2 × 2 x + 0 Here 6 x 3 is the dividend, 2 x is the divisor, 3 x 2 is the quotient and 0 is the remainder. Thus, in this way we can write any term in this format. i.e., Dividend = Quotient × Divisor + Remainder Hence, the first missing term is Quotient and then second is Divisor or vice versa.

E M B I B E

MATHEMATICS A TEXTBOOK FOR CLASS 8>Chapter 13 - Factorisation>Miscellaneous exercise

Jammu and Kashmir Board Solutions for Chapter: Factorisation, Exercise 8: Miscellaneous exercise

Author:Jammu and Kashmir Board

Jammu and Kashmir Board Mathematics Solutions for Exercise - Jammu and Kashmir Board Solutions for Chapter: Factorisation, Exercise 8: Miscellaneous exercise

Attempt the free practice questions on Chapter 13: Factorisation, Exercise 8: Miscellaneous exercise with hints and solutions to strengthen your understanding. MATHEMATICS A TEXTBOOK FOR CLASS 8 solutions are prepared by Experienced Embibe Experts.