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RS Aggarwal Solutions Class 9 Mathematics Solutions for Factorization of Polynomials Exercise 3C in Chapter 3 - Factorization of Polynomials

Question 64 Factorization of Polynomials Exercise 3C

(𝑎 + 𝑏)^2 + 101(𝑎 + 2𝑏) + 100

Answer:

(𝑎 + 𝑏)^2 + 101(𝑎 + 2𝑏) + 100

Consider,

𝑎 + 2𝑏 = x

So we get,

= (𝑥)^2 + 101^𝑥 + 100

We can further write it as

= (𝑥)^2 + 100𝑥 + 𝑥 + 100

By taking out the common terms

= x (x + 100) + 1 (x + 100)

= (x +1) (x + 100)

Let us replace x by a + 2b

So we get,

((a + 2b) +1) ((a + 2b) + 100)

= (a + 2b + 1) (a + 2b + 100)

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