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🔢 Chapter 4: Exploring Algebraic Identities • Exercise Set 4.2

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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 4.2, Q1: Factor completely:
(i) $9x^2 + 24xy + 16y^2$
(ii) $4s^2 + 20st + 25t^2$
(iii) $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$
(iv) $\frac{p^2}{16} - 2 + \frac{16}{p^2}$
(v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$

Answer:
(i) $(3x)^2 + 2(3x)(4y) + (4y)^2 = \mathbf{(3x + 4y)^2}$
(ii) $(2s)^2 + 2(2s)(5t) + (5t)^2 = \mathbf{(2s + 5t)^2}$
(iii) This is of the form $(a+b+c)^2$.
$\frac{m^2}{9} = (\frac{m}{3})^2$, $\frac{k^2}{4} = (\frac{k}{2})^2$, $9n^2 = (3n)^2$.
$2(\frac{m}{3})(\frac{k}{2}) = \frac{mk}{3}$, $2(\frac{m}{3})(3n) = 2mn$, $2(\frac{k}{2})(3n) = 3nk$.
Therefore, it factors to $\mathbf{(\frac{m}{3} + \frac{k}{2} + 3n)^2}$
(iv) $(\frac{p}{4})^2 - 2(\frac{p}{4})(\frac{4}{p}) + (\frac{4}{p})^2 = \mathbf{(\frac{p}{4} - \frac{4}{p})^2}$
(v) This is $(a+b+c)^2$. Since $-12ab$ and $-4bc$ are negative, but $6ac$ is positive, the term with $b$ must be negative. $(3a)^2 + (-2b)^2 + (c)^2 + 2(3a)(-2b) + 2(3a)(c) + 2(-2b)(c)$.
Factors to $\mathbf{(3a - 2b + c)^2}$

Q2

Exercise Set 4.2, Q2: Find the values of the following using the identity $(a-b)^2 = a^2 - 2ab + b^2$.
(i) $(79)^2$
(ii) $(193)^2$

Answer:
(i) $(80 - 1)^2 = 80^2 - 2(80)(1) + 1^2 = 6400 - 160 + 1 = \mathbf{6241}$
(ii) $(200 - 7)^2 = 200^2 - 2(200)(7) + 7^2 = 40000 - 2800 + 49 = \mathbf{37249}$