1. Find the product of the following binomials.
a. (5x + 3) (2x + 4)
\[ \begin{align*} &=(5x + 3)(2x + 4) \\ &= 5x \times (2x + 4) + 3 \times(2x + 4) \\ &= 5x \times 2x + 5x \times 4 + 3 \times 2x + 3 \times 4 \\ &= 10x^2 + 20x + 6x + 12 \\ &= 10x^2 + 26x + 12 \end{align*} \]
b. (7p - 3q) (2p + 5q)
\[ \begin{align*} &= (7p - 3q)(2p + 5q) \\ &= 7p \times (2p + 5p) - 3q \times (2p + 5q) \\ &= 7p \times 2p + 7p \times 5q - 3q \times 2p - 3q \times 5q \\ &= 14p^2 + 35pq - 6pq - 15q^2 \\ &= 14p^2 + 29pq - 15q^2 \end{align*} \]
c. (2⁄5 a + b) (a2 - 1⁄5 b2)
\[ \begin{align*} &= \left(\frac{2}{5}a + b\right) \times \left(a^2 - \frac{1}{5}b^2\right) \\ &= \frac{2}{5}a \times a^2 + \frac{2}{5}a \times \left(-\frac{1}{5}b^2\right) + b \times a^2 + b \times \left(-\frac{1}{5}b^2\right) \\ &= \frac{2}{5}a^3 - \frac{2}{25}ab^2 + a^2b - \frac{1}{5}b^3 \\ &= \frac{2}{5}a^3 - \frac{2}{25}ab^2 + a^2b - \frac{1}{5}b^3 \\ &= \frac{2}{5}a^3 + a^2b - \frac{2}{25}ab^2 - \frac{1}{5}b^3 \end{align*} \]
d. (m2n - 5) (6 - mn2)
\begin{align*} &=(m^2n - 5) \times (6 - mn^2) \\ & = m^2n \times (6 - mn^2) - 5 \times (6 - mn^2) \\ &= m^2n \times 6 - m^2n \times mn^2 - 5 \times 6 + 5 \times mn^2 \\ &= 6m^2n - m^3n^3 - 30 + 5mn^2 \\ &= 6m^2n + 5mn^2 - m^3n^3 - 30 \\ \end{align*}
e. (4x2 + 7y3) (3xy-2y2)
\begin{align*} &=(4x^2 + 7y^3) \times (3xy - 2y^2) \\ &= 4x^2 \times (3xy -2y^2) + 7y^3 \times (3xy - 2y^2) \\ &= 4x^2 \times 3xy + 4x^2 \times (-2y^2) + 7y^3 \times 3xy + 7y^3 \times (-2y^2) \\ &= 12x^3y - 8x^2y^2 + 21xy^4 - 14y^5 \\ \end{align*}
f. (1.1x + 2.7y) (1.1x - 2.7y)
\begin{align*} &=(1.1x + 2.7y) \times (1.1x - 2.7y) \\ &= 1.1x \times (1.1x - 2.7y) + 2.7y \times (1.1x - 2.7y) \\ &= 1.1x \times 1.1x + 1.1x \times (-2.7y) + 2.7y \times 1.1x + 2.7y \times (-2.7y) \\ &= 1.21x^2 - 2.97xy + 2.97xy - 7.29y^2 \\ &= 1.21x^2 - 7.29y^2 \\ \end{align*}
2. Multiply the following binomials and verify the results for the given values.
a. (2x2 - 5y) (5x + 2y2) ; x = 2, y = -1
\begin{align*} (2x^2 - 5y) \times (5x + 2y^2) \\ \text{Values: } & x = 2, y = -1 \\ \end{align*}
\begin{align*} &=(2x^2 - 5y) \times (5x + 2y^2) \\ &= 2x^2 \times (5x + 2y^2) - 5y \times (5x + 2y^2) \\ &= 2x^2 \times 5x + 2x^2 \times 2y^2 - 5y \times 5x - 5y \times 2y^2 \\ &= 10x^3 + 4x^2y^2 - 25xy - 10y^3 \\ \\ & \text{Verification} \\ & \text{L.H.S } \\ &=(2x^2 - 5y) \times (5x + 2y^2) \\ &=(2(2)^2 - 5(-1)) \times (5(2) + 2(-1)^2) \\ &=(2(4) + 5) \times (10 + 2) \\ &=(13) \times (12) \\ &= 156 \\ \\ & \text{R.H.S } \\ &= 10x^3 + 4x^2y^2 - 25xy - 10y^3 \\ &= 10(2)^3 + 4(2)^2(-1)^2 - 25(2)(-1) - 10(-1)^3 \\ &= 10 \times 8 + 4 \times 4 \times 1 - 25 \times 2 \times (-1) - 10 \times (-1) \\ &= 80 + 16 + 50 + 10 \\ &= 156 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}
b. (-1⁄4 a + 1⁄5b) (1⁄4 a + 1⁄5 b); a = 8, b = 5
\begin{align*} \left(-\frac{1}{4}a + \frac{1}{5}b\right) \times \left(\frac{1}{4}a + \frac{1}{5}b\right) \\ \text{Values: } & a = 8, b = 5 \\ \end{align*}
\begin{align*} &=\left(-\frac{1}{4}a + \frac{1}{5}b\right) \times \left(\frac{1}{4}a + \frac{1}{5}b\right) \\ &= -\frac{1}{4}a \times \left(\frac{1}{4}a + \frac{1}{5}b\right) + \frac{1}{5}b \times \left(\frac{1}{4}a + \frac{1}{5}b\right) \\ &= -\frac{1}{4}a \times \frac{1}{4}a - \frac{1}{4}a \times \frac{1}{5}b + \frac{1}{5}b \times \frac{1}{4}a + \frac{1}{5}b \times \frac{1}{5}b \\ &= -\frac{1}{16}a^2 - \frac{1}{20}ab + \frac{1}{20}ab + \frac{1}{25}b^2 \\ &= -\frac{1}{16}a^2 + \frac{1}{25}b^2 \\ \\ & \text{Verification} \\ & \text{L.H.S } \\ &=\left(-\frac{1}{4}a + \frac{1}{5}b\right) \times \left(\frac{1}{4}a + \frac{1}{5}b\right) \\ &=\left(-\frac{1}{4}(8) + \frac{1}{5}(5)\right) \times \left(\frac{1}{4}(8) + \frac{1}{5}(5)\right) \\ &=\left(-2 + 1\right) \times \left(2 + 1\right) \\ &= (-1) \times 3 \\ &= -3 \\ \\ & \text{R.H.S } \\ &= (-\frac{1}{16}a^2) + \frac{1}{25}b^2 \\ &= (-\frac{1}{16}(8)^2) + \frac{1}{25}(5)^2 \\ &= (-\frac{1}{16}) \times 64 + \frac{1}{25} \times 25 \\ &= (-4) + 1 \\ &= -3 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}
c. (m + n) (2m - 3n); m = -2; n = 0.
\begin{align*} (m + n) \times (2m - 3n) \\ \text{Values: } m = -2, n = 0 \\ \end{align*}
\begin{align*} &= (m + n) \times (2m - 3n) \\ &= m \times (2m - 3n) + n \times (2m - 3n) \\ &= m \times 2m + m \times (-3n) + n \times 2m + n \times (-3n) \\ &= 2m^2 - 3mn + 2mn - 3n^2 \\ &= 2m^2 - mn - 3n^2 \\ \\ & \text{Verification} \\ & \text{L.H.S} \\ &= (m + n) \times (2m - 3n) \\ &= (-2 + 0) \times (2(-2) - 3(0)) \\ &= (-2) \times (-4 - 0) \\ &= (-2) \times (-4) \\ &= 8 \\ \\ & \text{R.H.S} \\ &= 2m^2 - mn - 3n^2 \\ &= 2(-2)^2 - (-2)(0) - 3(0)^2 \\ &= 2 \times 4 - 0 - 0 \\ &= 8 - 0 - 0 \\ &= 8 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}
d. (0.1p + 0.2q) (0.2q - 0.1p); p = 10, q = 5.
\begin{align*} (0.1p + 0.2q) \times (0.2q - 0.1p) \\ \text{Values: } p = 10, q = 5 \\ \end{align*}
\begin{align*} &= (0.1p + 0.2q) \times (0.2q - 0.1p) \\ &= 0.1p \times (0.2q - 0.1p) + 0.2q \times (0.2q - 0.1p) \\ &= 0.1p \times 0.2q + 0.1p \times (-0.1p) + 0.2q \times 0.2q + 0.2q \times (-0.1p) \\ &= 0.02pq - 0.01p^2 + 0.04q^2 - 0.02pq \\ &= -0.01p^2 + 0.04q^2 \\ \\ & \text{Verification} \\ & \text{L.H.S} \\ &= (0.1p + 0.2q) \times (0.2q - 0.1p) \\ &= (0.1(10) + 0.2(5)) \times (0.2(5) - 0.1(10)) \\ &= (1 + 1) \times (1 - 1) \\ &= 2 \times 0 \\ &= 0 \\ \\ & \text{R.H.S} \\ &= -0.01p^2 + 0.04q^2 \\ &= -0.01(10)^2 + 0.04(5)^2 \\ &= -0.01 \times 100 + 0.04 \times 25 \\ &= -1 + 1 \\ &= 0 \\ \\ & \text{L.H.S = R.H.S} \\ & \text{Hence Verified} \\ \end{align*}
3. Simply the following
a. (2x - 3y) (3x + y) + (x + 2y) (x - y)
\begin{align*} [(2x - 3y) \times (3x + y)]+ [(x + 2y) \times (x - y)] \\ \end{align*}
\begin{align*} &= [(2x - 3y) \times (3x + y)]+ [(x + 2y) \times (x - y)] \\ &= [2x \times (3x + y) - 3y \times (3x + y)] + [x \times (x - y) + 2y \times (x - y)] \\ &= [2x \times 3x + 2x \times y - 3y \times 3x - 3y \times y] + [x \times x - x \times y + 2y \times x - 2y \times y] \\ &= [6x^2 + 2xy - 9xy - 3y^2] + [x^2 - xy + 2xy - 2y^2] \\ &= 6x^2 + x^2 + 2xy - 9xy - xy + 2xy - 3y^2 - 2y^2 \\ &= 7x^2 - 6xy - 5y^2 \end{align*}
b. (2⁄3 x + 4) (3⁄2 x + 6) - (1⁄7 x - 1) (1⁄7 x + 1)
\begin{align*} \left(\frac{2}{3} x + 4\right) \left(\frac{3}{2} x + 6\right) - \left(\frac{1}{7} x - 1\right) \left(\frac{1}{7} x + 1\right) \\ \end{align*}
\begin{align*} &= \left[\frac{2}{3} x \times \left(\frac{3}{2} x + 6\right) + 4 \times \left(\frac{3}{2} x + 6\right)\right] - \left[\frac{1}{7} x \times \left(\frac{1}{7} x + 1\right) - 1 \times \left(\frac{1}{7} x + 1\right)\right] \\ &= \left[\frac{2}{3} x \times \frac{3}{2} x + \frac{2}{3} x \times 6 + 4 \times \frac{3}{2} x + 4 \times 6\right] - \left[\frac{1}{49} x^2 + \frac{1}{7} x - \frac{1}{7} x - 1\right] \\ &= \left[1 \times x^2 + 4x + 6x + 24\right] - \left[\frac{1}{49} x^2 + 0 - 1\right] \\ &= x^2 + 10x + 24 - \frac{1}{49} x^2 + 1 \\ &= x^2 - \frac{1}{49} x^2 + 10x + 24 + 1 \\ &= \frac{48}{49} x^2 + 10x + 25 \\ &= \frac{48}{49} x^2 + 10x + 25 \end{align*}
c. (7 abc + b2) (7 - bc) + c(2 b3 - 9 ab)
\begin{align*} [(7 abc + b^2) (7 - bc)] + [c(2 b^3 - 9 ab)] \\ \end{align*}
\begin{align*} &= [(7 abc + b^2) (7 - bc)] + [c(2 b^3 - 9 ab)] \\ &= [7 abc \times (7 - bc) + b^2 \times (7 - bc)] + [c \times 2 b^3 - c \times 9 ab] \\ &= [7 abc \times 7 + 7 abc \times (-bc) + b^2 \times 7 + b^2 \times (-bc)] + [c \times 2 b^3 - c \times 9 ab] \\ &= 49 abc - 7 ab^2c^2 + 7 b^2 - b^3c + 2 b^3c - 9 abc \\ &= 49 abc - 9 abc - 7 ab^2c^2 + 7 b^2 - b^3c + 2 b^3c \\ &= 40 abc - 7 ab^2c^2 + 7 b^2 + b^3c \end{align*}
d. (p + q)(p2 - q2) + (p - q)(p2 + q2)
\begin{align*} [(p + q)(p^2 - q^2)] + [(p - q)(p^2 + q^2)] \\ \end{align*}
\begin{align*} &= [(p + q)(p^2 - q^2)] + [(p - q)(p^2 + q^2)] \\ &= [p(p^2 - q^2) + q(p^2 - q^2)] + [p(p^2 + q^2) - q(p^2 + q^2)] \\ &= [p^3 - pq^2 + p^2q - q^3] + [p^3 + pq^2 - p^2q - q^3] \\ &= p^3 - \cancel{pq^2} + \cancel{p^2q} - q^3 + p^3 + \cancel{pq^2} - \cancel{p^2q} - q^3 \\ &= p^3 + p^3 - q^3 - q^3 \\ &= 2p^3 - 2q^3 \end{align*}
e. (x + 2y)(-3x - y) - (x + y)(x - y) + (x - 2y)(-2x + y)
\begin{align*} [(x + 2y)(-3x - y)] - [(x + y)(x - y)] + [(x - 2y)(-2x + y)] \\ \end{align*}
\begin{align*} &= [(x + 2y)(-3x - y)] - [(x + y)(x - y)] + [(x - 2y)(-2x + y)] \\ &= [x(-3x - y) + 2y(-3x - y)] - [x(x - y) + y(x - y)] + [x(-2x + y) - 2y(2x - y)] \\ &= [-3x^2 - xy - 6xy - 2y^2] - [x^2 -\cancel{xy} + \cancel{xy} - y^2] + [-2x^2 + xy + 4xy - 2y^2] \\ &= [-3x^2 - 7xy - 2y^2] - [x^2 - y^2] + [-2x^2 + 5xy - 2y^2] \\ &= -3x^2 - 7xy - 2y^2 - x^2 + y^2 - 2x^2 + 5xy - 2y^2 \\ &= -3x^2 - x^2 - 2x^2 - 7xy + 5xy - 2y^2 + y^2 - 2y^2 \\ &= -6x^2 - 2xy - 3y^2 \end{align*}
f. (a2 + b2)(a2 + b2) - (a2 - b2)(a2 - b2)
\begin{align*} [(a^2 + b^2)(a^2 + b^2)] - [(a^2 - b^2)(a^2 - b^2)] \\ \end{align*}
\begin{align*} &= [(a^2 + b^2)(a^2 + b^2)] - [(a^2 - b^2)(a^2 - b^2)] \\ &= [a^2(a^2 + b^2) + b^2(a^2 + b^2)] - [a^2(a^2 - b^2) - b^2(a^2 - b^2)] \\ &= [a^4 + a^2b^2 + a^2b^2 + b^4] - [a^4 - a^2b^2 - a^2b^2 + b^4] \\ &= a^4 + 2a^2b^2 + b^4 - (a^4 - 2a^2b^2 + b^4) \\ &= \cancel{a^4} + 2a^2b^2 + \cancel{b^4} - \cancel{a^4} + 2a^2b^2 - \cancel{b^4} \\ &= 2a^2b^2 + 2a^2b^2 \\ &= 4a^2b^2 \end{align*}