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End-of-Chapter Exercises, Q1: Use suitable identities to find the following products:
(i) $(-3x + 4)^2$
(ii) $(2s + 7)(2s - 7)$
(iii) $(p^2 + \frac{1}{2})(p^2 - \frac{1}{2})$
(v) $(s - 2t)(s^2 + 2st + 4t^2)$
(vi) $(\frac{1}{2r} - 4r)^2$
(viii) $(x - \frac{1}{3}y)^3$
Answer:
(i) $(-3x)^2 + 2(-3x)(4) + 4^2 = \mathbf{9x^2 - 24x + 16}$
(ii) $(2s)^2 - 7^2 = \mathbf{4s^2 - 49}$
(iii) $(p^2)^2 - (\frac{1}{2})^2 = \mathbf{p^4 - \frac{1}{4}}$
(v) This is $(a-b)(a^2+ab+b^2) = a^3 - b^3$. So, $s^3 - (2t)^3 = \mathbf{s^3 - 8t^3}$
(vi) $(\frac{1}{2r})^2 - 2(\frac{1}{2r})(4r) + (4r)^2 = \mathbf{\frac{1}{4r^2} - 4 + 16r^2}$
(viii) $x^3 - 3(x^2)(\frac{1}{3}y) + 3(x)(\frac{1}{9}y^2) - (\frac{1}{3}y)^3 = \mathbf{x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3}$
End-of-Chapter Exercises, Q4: Simplify the following:
(i) $\frac{4x^2 + 4x + 1}{4x^2 - 1}$
(ii) $\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}$
(iii) $\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}$
Answer:
(i) $\frac{(2x+1)^2}{(2x-1)(2x+1)} = \mathbf{\frac{2x+1}{2x-1}}$
(ii) $\frac{27(a^3 - 8b^3)}{9(a^2 - 4b^2)} = \frac{3(a-2b)(a^2+2ab+4b^2)}{(a-2b)(a+2b)} = \mathbf{\frac{3(a^2+2ab+4b^2)}{a+2b}}$
(iii) $\frac{(s+5t)(s^2-5st+25t^2)}{(s-7t)(s+5t)} = \mathbf{\frac{s^2-5st+25t^2}{s-7t}}$
End-of-Chapter Exercises, Q5: Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) $25a^2 - 30ab + 9b^2$
(ii) $36s^2 - 49t^2$
Answer:
(i) This factors to $(5a - 3b)^2$. Length and breadth can both be $\mathbf{5a - 3b}$ (making it a square).
(ii) This is a difference of squares: $(6s)^2 - (7t)^2$. Length and breadth are $\mathbf{(6s - 7t)}$ and $\mathbf{(6s + 7t)}$.
End-of-Chapter Exercises, Q6: Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) $6a^2 - 24b^2$
(ii) $3ps^2 - 15ps + 12p$
Answer:
(i) $6(a^2 - 4b^2) = 6(a - 2b)(a + 2b)$. Dimensions: $\mathbf{6}$, $\mathbf{(a - 2b)}$, and $\mathbf{(a + 2b)}$.
(ii) $3p(s^2 - 5s + 4) = 3p(s - 1)(s - 4)$. Dimensions: $\mathbf{3p}$, $\mathbf{(s - 1)}$, and $\mathbf{(s - 4)}$.
End-of-Chapter Exercises, Q7: The village playground is shaped as a square of side 40 metres. A path of width $s$ metres is created around the playground for people to walk. Find an expression for the area of the path in terms of $s$.
Answer: The outer dimensions of the path + playground will be $(40 + 2s)$.
The area of the path is the outer square area minus the playground area.
Area = $(40 + 2s)^2 - 40^2 = (40 + 2s - 40)(40 + 2s + 40) = 2s(80 + 2s) = \mathbf{160s + 4s^2}$ sq. meters.
End-of-Chapter Exercises, Q11: If $a + b + c = 5$ and $ab + bc + ca = 10$, then prove that $a^3 + b^3 + c^3 - 3abc = -25$.
Answer: We know the identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))$.
First find $a^2+b^2+c^2$.
$(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$.
$5^2 = a^2+b^2+c^2 + 2(10) \implies 25 = a^2+b^2+c^2 + 20 \implies a^2+b^2+c^2 = 5$.
Substitute back into the main identity:
$a^3 + b^3 + c^3 - 3abc = (5)(5 - 10) = 5(-5) = \mathbf{-25}$.
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