Solution

\[ \begin{align*} &= (3p^2 - 14pq + 2r) - (14pq + 3p^2 + 2r^2) \\ &= 3p^2 - 14pq + 2r - 14pq - 3p^2 - 2r^2 \\ &= \cancel{3p^2} - \cancel{3p^2} - 14pq - 14pq + 2r - 2r^2 \\ &= -28pq + 2r - 2r^2 \end{align*} \]

Answer \( \color{red} (iii)\text{ trinomial} \)

Solution

\[ \begin{align*} \text{Terms: } & 4p^3q^2r , - 12pq^2r^2, 16p^2q^2r^2 \\ HCF & = 4pq^2r \end{align*} \]

Answer \( \ \color{red}(i) \ 4pq^2r \)

Solution

\[ \begin{align*} 2x^2 - 5x + 10 & = 2x^2 - tx + 2t \\ \text{Comparing }& \text{the terms} \\ t & = 5 \\ \end{align*} \]

Answer \( \ \color{red}(ii) \ 5 \)

Solution

\[ \begin{align*} z^2 + 3z - p = 3 \ &; \ z = 2 \\ (2)^2 + 3(2) - p &= 3 \\ 4 + 6 - p &= 3 \\ 10 - p &= 3 \\ -p &= 3 - 10 \\ - p &= - 7 \\ p &= 7 \\ \end{align*} \]

Answer \( \ \color{red}(iii) \ 7 \)

Solution

\[ \begin{align*} & = 2x^3 - 6x^2 + 5x - 15 \\ & = 2x^2(x - 3) + 5(x - 3) \\ & = (2x^2 + 5)(x - 3) \end{align*} \]

Answer \( \ \color{red}(iii) \ (2x^2 + 5) \text{ and } (x - 3) \)

Solution

\[ \begin{align*} \text{Draw PQ} & \text{ perpendicular to AB} \\ Base & = p^2q - 5 \\ Height & = q^2 + 4p \\ \\ \text{Area of } \triangle PAB & = \frac{1}{2} \times Base \times Height \\ \\ & = \frac{1}{2} [(p^2q - 5) \times (q^2 + 4p)]\\ \\ & = \frac{1}{2} [p^2q(q^2 + 4p) - 5(q^2 + 4p)] \\ \\ & = \frac{1}{2} (p^2q^3 + 4p^3q - 5q^2 - 20p) \text{ sq. units} \\ \\ \end{align*} \]

Answer Area of \( \triangle PAB = \color{red} \frac{1}{2} (p^2q^3 + 4p^3q - 5q^2 - 20p) \text{ sq. units} \)

Solution

\[ \begin{align*} & = (72z^2 - 45z + 4) - (30z-42z^2-17) \\ & = 72z^2 - 45z + 4 - 30z + 42z^2 + 17 \\ & = \color{magenta} 72z^2 + 42z^2 \color{green} - 45z - 30z \color{blue} + 4 + 17 \\ & = \color{magenta} 114z^2 \color{green} - 75z \color{blue} + 21 \\ \end{align*} \]

Answer Exceeds the expression by \( \color{red} (114z^2 - 75z + 21) \)

Solution

\[ \begin{align*} \text{Perimeter of triangle} &= x^2y + 10 \\ \text{Side 1 + Side 2 + Side 3 } & = \text{Perimeter of triangle} \\ (x^2y - 4) + (3 - 2x^2y) + \text{(Side 3) } & = x^2y + 10 \\ x^2y - 4 + 3 - 2x^2y + \text{(Side 3) } & = x^2y + 10 \\ -x^2y - 1 + \text{(Side 3) } & = x^2y + 10 \\ \text{Side 3 } & = x^2y + 10 + x^2y + 1 \\ \text{Side 3} &= (2x^2y + 11) \ units \end{align*} \]

Answer Length of third side \( = \color{red} (2x^2y + 11) \ units\)

Solution

\[ \begin{align*} 2b^3c &= \boxed{2} \times \boxed{b} \times \boxed{b} \times b \times \boxed{c} \\ 4b^4c^2 &= \boxed{2} \times 2 \times \boxed{b} \times \boxed{b} \times b \times b \times \boxed{c} \times c \\ 16b^2c^2 &= \boxed{2} \times 2 \times 2 \times 2 \times \boxed{b} \times \boxed{b} \times \boxed{c} \times c \\ \text{HCF} &= 2b^2c \\ \\ & = 2b^3c + 4b^4c^2 + 16b^2c^2 \\ & = 2b^2c(b + 2b^2c+8c) \\ Factors & = 2b^2c \ and \ b + 2b^2c+8c \end{align*} \]

Answer HCF \( = \color{red} 2b^2c \) , Factors \( = \color{red} 2b^2c \ , \ b + 2b^2c+8c\)

Solution

\[ \begin{align*} Length & = a - b \\ Breadth & = 2a + b \\ Area & = Length \times Breadth \\ & = (a - b) (2a + b) \\ & = a(2a + b) -b (2a + b) \\ & = 2a^2+ab-2ab-b^2 \\ & = 2a^2-1ab-b^2 \\ \\ &Given \\ 2a^2-1ab-b^2 & = Kab + 2a^2 - b^2 \\ \\ \text{By comparing } & \text{the coefficients of term 'ab'} \\ K & = -1 \end{align*} \]

Answer \( K = \color{red} -1 \)

Solution

\[ \begin{align*} & = (13ax - 4)(5ay + 1) \\ & = 13ax(5ay + 1) - 4(5ay + 1) \\ & = 65a^2xy + 13ax - 20ay - 4 \end{align*} \]

Solution

\[ \begin{align*} & = (3x^2 + 5x - 7)(x + 5y) \\ & = 3x^2(x + 5y) + 5x(x + 5y) - 7(x + 5y) \\ & = 3x^3 + 15x^2y + 5x^2 + 25xy - 7x - 35y \end{align*} \]

Solution

\[ \begin{align*} & = -6a^2b(a^4 + b^4 - 3a^2b^2) \\ & = -6a^6b - 6a^2b^5 + 18a^4b^3 \end{align*} \]

Solution

\[ \begin{align*} & = \left(\frac{3}{5}x - \frac{2}{9}y \right)(15x - 9y) \\ \\ & = \frac{3}{5}x(15x - 9y) - \frac{2}{9}y(15x - 9y) \\ \\ & = \left(\frac{3}{\cancel5}x \times \cancel{15}^3x \right) -\left(\frac{3}{5}x \times 9y \right) - \left( \frac{2}{\cancel9_3}y \times \cancel{15}^5x \right) + \left( \frac{2}{\cancel9}y \times \cancel9 y \right) \\ \\ & = 9x^2 - \frac{27}{5}xy - \frac{10}{3}xy + 2y^2 \\ \\ & = 9x^2 - \frac{81 - 50}{15}xy + 2y^2 \\ \\ & = 9x^2 - \frac{131}{15}xy + 2y^2 \end{align*} \]

Solution

\[ \begin{align*} & = (x^9)(-x^{10})(x^{11})(-x^{12}) \\ & = x^{42} \end{align*} \]

Solution

\[ \begin{align*} & = 0.9p^3q^3(10p - 20q) \\ & = (0.9 \times 10)p^4q^3 - (0.9 \times 20)p^3q^4 \\ & = 9p^4q^3 - 18p^3q^4 \end{align*} \]

Solution

\[ \begin{align*} & = (7x^2 - 11x + 10)(x^3 - x^2) \\ & = 7x^2(x^3 - x^2) - 11x(x^3 - x^2) + 10(x^3 - x^2) \\ & = 7x^5 - 7x^4 - 11x^4 + 11x^3 + 10x^3 - 10x^2 \\ & = 7x^5 - 18x^4 + 21x^3 - 10x^2 \\ \end{align*} \]

Solution

\[ \begin{align*} & = \left(\frac{10}{9}a^5b^6c^6\right)\left(-\frac{3}{2}b^2c\right)\left(\frac{6}{5}c^3d^4\right)(-a^3d^5) \\ \\ & = \left(\frac{10}{9} \times -\frac{3}{2} \times \frac{6}{5} \times -1 \right) \times a^5 \times a^3 \times b^6 \times b^2 \times c^6 \times c^1 \times c^3 \times d^4 \times d^5 \\ \\ & = \frac{180}{90}a^8b^8c^{10}d^9 \\ \\ & = 2a^8b^8c^{10}d^9 \end{align*} \]

Solution

\[ \begin{align*} & = (0.7x^3 - 0.5y^3)(0.7x^3 + 0.5y^3) \\ & = 0.7x^3(0.7x^3 + 0.5y^3) - 0.5y^3(0.7x^3 + 0.5y^3) \\ & = 0.49x^6 + \cancel{0.35x^3y^3} - \cancel{0.35x^3y^3} - 0.25y^6 \\ & = 0.49x^6 - 0.25y^6 \\ \end{align*} \]

Solution

\[ \begin{align*} & = (64a^2 - 56ab + 49b^2)(8a + 7b) \\ & = 64a^2(8a + 7b) - 56ab(8a + 7b) + 49b^2 (8a + 7b) \\ & = 512a^3 + \cancel{448a^2b} - \cancel{448a^2b} - \cancel{392ab^2} + \cancel{392ab^2} + 343b^3 \\ & = 512a^3 + 343b^3 \\ \end{align*} \]

Solution

\[ \begin{align*} \text{Variable part }& = (x^3y^9) (x^6y^5) (y^2z^4) \\ & = (x^3 \times x^6) \times (y^9 \times y^5 \times y^2) \times z^4 \\ & = x^9 \times y^{16} \times z^4 \\ & = x^9 y^{16} z^4 \end{align*} \]

Solution

\[ \begin{align*} \text{Variable part }& = (a^2b^4) (b^3) (a^6b) \\ & = (a^2 \times a^6) \times (b^4 \times b^3 \times b^1) \\ & = a^8 \times b^8 \\ & = a^8 b^8 \end{align*} \]

Solution

\[ \begin{align*} & = (5x^6)(12x^2y)\left(\frac{3}{20}xy^2\right) \\ \\ & = \left(\cancel{5}^1 \times \cancel{12}^3 \times \frac{3}{{\cancel{20}_\cancel{4}}_1}\right) \times (x^6 \times x^2 \times x) \times (y \times y^2) \\ \\ & = 9 \times x^{9} \times y^{3} \\ & = 9x^9y^3 \\ \\ & \text{Evaluate for } x = 1, y = 2 \\ \\ & = 9x^9y^3 \\ & = 9(1)^9(2)^3 \\ & = 9 \times 1 \times 8 \\ & = 72 \end{align*} \]

Answer \( \color{red} 9x^9y^3 \) and \( \color{red} 72 \)

Solution

\[ \begin{align*} & = 1.5a^2(10ab - 4b^2) \\ & = (1.5a^2 \times 10ab) - (1.5a^2 \times 4b^2) \\ & = 15a^3b - 6a^2b^2 \\ \\ &\text{Evaluate for } a = -2, b = 3 \\ \\ & = 15a^3b - 6a^2b^2 \\ & = 15(-2)^3(3) - 6(-2)^2(3)^2 \\ & = 15(-8)(3) - 6(4)(9) \\ & = [15 \times (-24)] - (6 \times 36)\\ & = -360 - 216 \\ & = -576 \end{align*} \]

Answer \( \color{red} 15a^3b - 6a^2b^2 \) and \( \color{red} -576 \)

\[ \begin{align*} & = (3x - 4y)(4x^2y + 3xy^2) \\ & = 3x (4x^2y + 3xy^2) - 4y (4x^2y + 3xy^2) \\ & = 12x^3y + 9x^2y^2 - 16x^2y^2 - 12xy^3 \\ & = 12x^3y - 7x^2y^2 - 12xy^3 \\ \\ & {\color{magenta} \text{Verification: }} x = 2, y = -1 \\ & \color{magenta} \text{L.H.S} \\ & = (3x - 4y)(4x^2y + 3xy^2) \\ & = [3(2) - 4(-1)] \times [4(2^2)(-1) + 3(2)(-1)^2] \\ & = (6 + 4) \times [(-16) + 6] \\ & = 10 \times (-10) \\ & = -100 \\ \\ & \color{magenta} \text{R.H.S} \\ & = 12x^3y - 7x^2y^2 - 12xy^3 \\ & = 12(2^3)(-1) - 7(2^2)(-1)^2 - 12(2)(-1)^3 \\ & = 12(8)(-1) - 7(4)(1) - 12(2)(-1) \\ & = -96 - 28 + 24 \\ & = -100 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*} \]

\[ \begin{align*} & = \left( \frac{1}{4}a^2 + \frac{5}{9}b^2 \right)(a + b + ab) \\ \\ & = \frac{1}{4}a^2(a + b + ab) + \frac{5}{9}b^2(a + b + ab) \\ \\ & = \frac{1}{4}a^3 + \frac{1}{4}a^2b + \frac{1}{4}a^3b + \frac{5}{9}ab^2 + \frac{5}{9}b^3 + \frac{5}{9}ab^3 \\ \\ & {\color{magenta} \text{Verification: }} a = 2, b = 3 \\ & \color{magenta} \text{L.H.S} \\ & = \left( \frac{1}{4}a^2 + \frac{5}{9}b^2 \right)(a + b + ab) \\ & = \left( \frac{1}{4}(2^2) + \frac{5}{9}(3^2) \right) \times (2 + 3 + 2(3)) \\ & = \left( \frac{1}{\cancel4} \times \cancel4 + \frac{5}{\cancel9}\times \cancel9 \right) \times (2 + 3 + 6) \\ & = \left( 1 + 5 \right) \times 11 \\ & = 6 \times 11 \\ & = 66 \\ \\ & \color{magenta} \text{R.H.S} \\ & = \frac{1}{4}a^3 + \frac{1}{4}a^2b + \frac{1}{4}a^3b + \frac{5}{9}ab^2 + \frac{5}{9}b^3 + \frac{5}{9}ab^3 \\ \\ & = \frac{1}{4}(2)^3 + \frac{1}{4}{(2)}^2(3) + \frac{1}{4}{(2)}^3(3) + \frac{5}{9}(2)(3)^2 + \frac{5}{9}(3)^3 + \frac{5}{9}(2)(3)^3 \\ \\ & = \left( \frac{1}{\cancel4} \times \cancel{8}^2 \right) + \left( \frac{1}{\cancel4} \times \cancel4 \times 3 \right) + \left( \frac{1}{\cancel4} \times \cancel8^2 \times 3 \right) + \left( \frac{5}{\cancel9} \times 2 \times \cancel9 \right) + \left( \frac{5}{\cancel9} \times \cancel{27}^3 \right) + \left( \frac{5}{\cancel9} \times 2 \times \cancel{27}^3 \right) \\ \\ & = 2 + 3 + 6 + 10 + 15 + 30 \\ & = 66 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*} \]

\[ \begin{align*} & = (x^3y - y^2)(x^3y + y^2) \\ & = x^3y(x^3y + y^2) - y^2(x^3y + y^2) \\ & = x^6y^2 + \cancel{x^3y^3} - \cancel{x^3y^3} - y^4 \\ & = x^6y^2 - y^4 \\ \\ & {\color{magenta} \text{Verification: }} x = -1, y = -2 \\ & \color{magenta} \text{L.H.S} \\ & = (x^3y - y^2)(x^3y + y^2) \\ & = [(-1)^3(-2) - (-2)^2][(-1)^3(-2) + (-2)^2] \\ & = [ (-1) (-2) - 4][(-1)(-2) + 4] \\ & = [ 2 - 4][2 + 4] \\ & = -2 \times 6 \\ & = -12 \\ \\ & \color{magenta} \text{R.H.S} \\ & = x^6y^2 - y^4 \\ & = (-1)^6(-2)^2 - (-2)^4 \\ & = (1 \times 4) - 16 \\ & = 4 - 16 \\ & = -12 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*} \]

\[ \begin{align*} & = (2p + 3q)(4p^2 + 12pq + 9q^2) \\ & = 2p(4p^2 + 12pq + 9q^2) + 3q(4p^2 + 12pq + 9q^2) \\ & = 8p^3 + 24p^2q + 18pq^2 + 12p^2q + 36pq^2 + 27q^3 \\ & = 8p^3 \color{magenta} + 24p^2q + 12p^2q \color{green} + 18pq^2 + 36pq^2 \color{black}+ 27q^3 + 27q^3 \\ & = 8p^3 \color{magenta} + 36p^2q \color{green} + 54pq^2 \color{black}+ 27q^3 \\ \\ & {\color{magenta} \text{Verification: }} p = \frac{1}{2}, q = \frac{1}{3} \\ & \color{magenta} \text{L.H.S} \\ & = (2p + 3q)(4p^2 + 12pq + 9q^2) \\ \\ & = \left[ \cancel2 \left( \frac{1}{\cancel2} \right) + \cancel3\left( \frac{1}{\cancel3} \right) \right] \left[ 4\left( \frac{1}{2}\right)^2 + 12\left( \frac{1}{2} \right)\left( \frac{1}{3} \right) + 9\left( \frac{1}{3} \right)^2 \right] \\ \\ & = ( 1 + 1 ) \left[ \cancel4 \left( \frac{1}{\cancel4}\right) + \cancel{12}^2 \left( \frac{1}{\cancel 6} \right) + \cancel9\left( \frac{1}{\cancel9} \right) \right] \\ \\ & = (2)(1 + 2 + 1) \\ & = 2 \times 4 \\ & = 8 \\ \\ & \color{magenta} \text{R.H.S} \\ & = 8p^3 + 36p^2q + 54pq^2 + 27q^3 \\ \\ & = 8\left( \frac{1}{2} \right)^3 + 36\left( \frac{1}{2} \right)^2\left( \frac{1}{3} \right) + 54\left( \frac{1}{2} \right)\left( \frac{1}{3} \right)^2 + 27\left( \frac{1}{3} \right)^3 \\ \\ & = \cancel8 \left( \frac{1}{\cancel8} \right) + \cancel{36}^3 \left( \frac{1}{\cancel{12}} \right) + \cancel{54}^3 \left( \frac{1}{\cancel{18}} \right) + \cancel{27} \left( \frac{1}{\cancel{27}} \right) \\ \\ & = 1 + 3 + 3 + 1 \\ & = 8 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*} \]

\[ \begin{align*} & = (m^2 + mn + n^2)(m - n) \\ & = m^2(m - n) + mn(m - n) + n^2(m - n) \\ & = m^3 - \cancel{m^2n} + \cancel{m^2n} - \cancel{mn^2} + \cancel{mn^2} - n^3 \\ & = m^3 - n^3 \\ \\ & {\color{magenta} \text{Verification: }} m = 4, n = 3 \\ & \color{magenta} \text{L.H.S} \\ & = (m^2 + mn + n^2)(m - n) \\ & = [(4)^2 + (4)(3) + (3)^2] \times (4 - 3) \\ & = [16 + 12 + 9] \times 1 \\ & = 37 \\ \\ & \color{magenta} \text{R.H.S} \\ & = m^3 - n^3 \\ & = (4)^3 - (3)^3 \\ & = 64 - 27 \\ & = 37 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*} \]

\[ \begin{align*} & = 3x^2(3y^2 + 2) - x(x - 2xy^2) + y(2x^2y - 2y) \\ & = 9x^2y^2 + 6x^2 - x^2 + 2x^2y^2 + 2x^2y^2 - 2y^2 \\ & = \color{magenta} 9x^2y^2 + 2x^2y^2 + 2x^2y^2 \color{green} + 6x^2 - x^2 \color{black} - 2y^2 \\ & = \color{magenta} 13x^2y^2 \color{green} + 5x^2 \color{black} - 2y^2 \end{align*} \]

Answer \( \color{red} 13x^2y^2 + 5x^2 - 2y^2 \)

\[ \begin{align*} & = (2x + 7)(5x + 9) - (4x^2 + 1)(x - 3) \\ & = [2x (5x + 9) +7(5x + 9)] - [4x^2(x - 3) +1(x - 3)] \\ & = [10x^2 + 18x + 35x + 63] - [4x^3 - 12x^2 + x - 3] \\ & = 10x^2 + 53x + 63 - 4x^3 + 12x^2 - x + 3 \\ & = - 4x^3 \color{magenta}+ 10x^2 + 12x^2 \color{green}- x + 53x \color{blue}+ 63 + 3 \\ & = -4x^3 \color{magenta} + 22x^2 \color{green} + 52x \color{blue} + 66 \end{align*} \]

Answer \( \color{red} -4x^3 + 22x^2 + 52x + 66 \)

\[ \begin{align*} & = (y^2 - 7y + 4)(3y^2 - 2) + (y + 1)(y^2 + 2y) \\ & = y^2(3y^2 - 2) - 7y(3y^2 - 2) + 4(3y^2 - 2) + y(y^2 + 2y) + 1(y^2 + 2y) \\ & = 3y^4 - 2y^2 - 21y^3 + 14y + 12y^2 - 8 + y^3 + 2y^2 + y^2 + 2y \\ & = 3y^4 \color{magenta} - 21y^3 + y^3 \color{green} - 2y^2 + 12y^2 + 2y^2 + y^2 \color{blue} + 2y + 14y \color{black} - 8 \\ & = 3y^4 \color{magenta} - 20y^3 \color{green} + 13y^2 \color{blue} + 16y \color{black} - 8 \end{align*} \]

Answer \( \color{red} 3y^4 - 20y^3 + 13y^2 + 16y - 8 \)

Solution

\[ \begin{align*} 3a^2b^2 &= \boxed{3} \times \boxed{a} \times a \times \boxed{b} \times \boxed{b} \\ 6ab^2c^2 &= \boxed{3} \times 2 \times \boxed{a} \times \boxed{b} \times \boxed{b} \times c \times c \\ 12a^2b^2c^2 &= \boxed{3} \times 2 \times 2 \times \boxed{a} \times a \times \boxed{b} \times \boxed{b} \times c \times c \\ \\ \text{HCF} &= 3 \times a \times b \times b \\ &= 3ab^2 \end{align*} \]

Answer HCF \( = \color{red} 3ab^2 \)

Solution

\[ \begin{align*} & = ab^2 - ab - bc^2 + c^2 \\ & = ab(b - 1) - c^2(b - 1) \\ & = (b - 1)(ab - c^2) \end{align*} \]

Answer \( \color{red} (b - 1)(ab - c^2) \)

Solution

\[ \begin{align*} & = 4(p + q)(3a - b) + 6(p + q)(2b - 3a) \\ & = 2(p + q) \big[2(3a - b) + 3(2b - 3a)\big] \\ & = 2(p + q) \big[6a - 2b + 6b - 9a\big] \\ & = 2(p + q) \big[6a - 9a - 2b + 6b - 9a\big] \\ & = 2(p + q)(-3a + 4b) \end{align*} \]

Answer \( \color{red} 2(p + q)(-3a + 4b) \)

Solution

\[ \begin{align*} & = axy + bcxy - az - bcz \\ & = xy(a + bc) - z(a + bc) \\ & = (a + bc)(xy-z) \end{align*} \]

Answer \( \color{red} (a + bc)(xy-z) \)