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🔢 Chapter 4: Exploring Algebraic Identities • Exercise Set 4.4

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Exercise Set 4.1 2Exercise Set 4.2 2Exercise Set 4.3 1Exercise Set 4.4 2Exercise Set 4.5 1End-of-Chapter Exercises 6All Exercises 14
Q1

Exercise Set 4.4, Q1: Fill in the blanks to complete the following identities:
(i) $s^2 - 11s + 24 = (\_\_)(\_\_)$
(ii) $(\_\_)(x + 1) = (3x^2 - 4x - 7)$
(iii) $10x^2 - 11x - 6 = (2x - \_\_)(\_\_ + 2)$
(iv) $6x^2 + 7x + 2 = (\_\_)(\_\_)$

Answer:
(i) Find factors of 24 that sum to -11 (-8 and -3). $\mathbf{(s - 8)(s - 3)}$
(ii) Divide $(3x^2 - 4x - 7)$ by $(x + 1)$. The missing factor is $\mathbf{(3x - 7)}$.
(iii) $10x^2 - 11x - 6 = (2x - 3)(5x + 2)$. So the blanks are $\mathbf{3}$ and $\mathbf{5x}$.
(iv) Splitting the middle term: $6x^2 + 4x + 3x + 2 = 2x(3x + 2) + 1(3x + 2)$. So, $\mathbf{(2x + 1)(3x + 2)}$.

Q2

Exercise Set 4.4, Q2: Select and use the identity that will help you to find the following products without multiplying directly:
(i) $(41)^2$
(ii) $(27)^2$
(iii) $(23 \times 17)$
(iv) $(135)^2$
(v) $(97)^2$
(vi) $(18 \times 29)$
(vii) $(34 \times 43)$
(viii) $(205)^2$

Answer:
(i) $(40 + 1)^2 = 1600 + 80 + 1 = \mathbf{1681}$
(ii) $(30 - 3)^2 = 900 - 180 + 9 = \mathbf{729}$
(iii) $(20 + 3)(20 - 3) = 20^2 - 3^2 = 400 - 9 = \mathbf{391}$
(iv) $(130 + 5)^2 = 16900 + 1300 + 25 = \mathbf{18225}$
(v) $(100 - 3)^2 = 10000 - 600 + 9 = \mathbf{9409}$
(vi) $(20 - 2)(30 - 1) = 600 - 20 - 60 + 2 = \mathbf{522}$ (Using $(x+a)(x+b)$ or distributive)
(vii) $(40 - 6)(40 + 3) = 1600 + 120 - 240 - 18 = \mathbf{1462}$
(viii) $(200 + 5)^2 = 40000 + 2000 + 25 = \mathbf{42025}$