RD Sharma solution class 8 chapter 6 Algebraic Expressions and Identity Exercise 6.7

Exercise 6.7

Page-6.47

Question 1:

Find the following products:
(i) (x + 4) (x + 7)
(ii) (x − 11) (x + 4)
(iii) (x + 7) (x − 5)
(iv) (x − 3) ( x − 2)
(v) (y² − 4) (y² − 3)
(vi) $\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)$
(vii) (3x + 5) (3x + 11)
(viii) (2x² − 3) (2x² + 5)
(ix) (z² + 2) (z² − 3)
(x) (3x − 4y) (2x − 4y)
(xi) (3x² − 4xy) (3x² − 3xy)
(xii) $\left(x+\frac{1}{5}\right)(x + 5)$
(xiii) $\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)$
(xiv) (x² + 4) (x² + 9)
(xv) (y² + 12) (y² + 6)
(xvi) $\left(y^2+\frac{5}{7}\right)\left(y^2-\frac{14}{5}\right)$
(xvii) (p² + 16) $\left(p^2-\frac{1}{4}\right)$

Answer 1:

(i) Here, we will use the identity $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab.$
$$\begin{gathered} \left(x+4\right)\left(x+7\right) \\ =x^2+\left(4+7\right)x+4\times 7 \\ =x^2+11x+28 \end{gathered}$$

(ii) Here, we will use the identity $\left(x-a\right)\left(x+b\right)=x^2+\left(b-a\right)x-ab$.
$$\begin{gathered} \left(x-11\right)\left(x+4\right) \\ =x^2+\left(4-11\right)x-11\times 4 \\ =x^2-7x-44 \end{gathered}$$

(iii) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)=x^2+\left(a-b\right)x-ab$.
$$\begin{gathered} \left(x+7\right)\left(x-5\right) \\ =x^2+\left(7-5\right)x-7\times 5 \\ =x^2+2x-35 \end{gathered}$$

(iv) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)=x^2-\left(a+b\right)x+ab$.
$$\begin{gathered} \left(x-3\right)\left(x-2\right) \\ =x^2-\left(3+2\right)x+3\times 2 \\ =x^2-5x+6 \end{gathered}$$

(v) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)=x^2-\left(a+b\right)x+ab$.
$$\begin{gathered} \left(y^2-4\right)\left(y^2-3\right) \\ ={\left(y^2\right)}^2-\left(4+3\right)\left(y^2\right)+4\times 3 \\ =y^4-7y^2+12 \end{gathered}$$

(vi) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$.
$$\begin{gathered} \left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right) \\ =x^2+\left(\frac{4}{3}+\frac{3}{4}\right)x+\frac{4}{3}\times \frac{3}{4} \\ =x^2+\frac{25}{12}x+1 \end{gathered}$$

(vii) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$.
$$\begin{gathered} \left(3x+5\right)\left(3x+11\right) \\ ={\left(3x\right)}^2+\left(5+11\right)\left(3x\right)+5\times 11 \\ =9x^2+48x+55 \end{gathered}$$

(viii) Here, we will use the identity​ $\left(x-a\right)\left(x+b\right)=x^2+\left(b-a\right)x-ab$.
$$\begin{gathered} \left(2x^2-3\right)\left(2x^2+5\right) \\ ={\left(2x^2\right)}^2+\left(5-3\right)\left(2x^2\right)-3\times 5 \\ =4x^4+4x^2-15 \end{gathered}$$

(ix) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)=x^2+\left(a-b\right)x-ab$.
$$\begin{gathered} \left(z^2+2\right)\left(z^2-3\right) \\ ={\left(z^2\right)}^2+\left(2-3\right)\left(z^2\right)-2\times 3 \\ =z^4-z^2-6 \end{gathered}$$

(x) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)=x^2-\left(a+b\right)x+ab$.
$$\begin{gathered} \left(3x-4y\right)\left(2x-4y\right) \\ =\left(4y-3x\right)\left(4y-2x\right) (\mathrm{Taking} \mathrm{common} -1 \mathrm{from} \mathrm{both} \mathrm{parentheses}) \\ ={\left(4y\right)}^2-\left(3x+2x\right)\left(4y\right)+3x\times 2x \\ =16y^2-\left(12xy+8xy\right)+6x^2 \\ =16y^2-20xy+6x^2 \end{gathered}$$

(xi) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)=x^2-\left(a+b\right)x+ab$.

$$\begin{gathered} \left(3x^2-4xy\right)\left(3x^2-3xy\right) \\ ={\left(3x^2\right)}^2-\left(4xy+3xy\right)\left(3x^2\right)+4xy\times 3xy \\ =9x^4-\left(12x^{3}y+9x^{3}y\right)+12x^2{y}^2 \\ =9x^4-21x^{3}y+12x^2{y}^2 \end{gathered}$$

(xii) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$.
$$\begin{gathered} \left(x+\frac{1}{5}\right)\left(x+5\right) \\ =x^2+\left(\frac{1}{5}+5\right)x+\frac{1}{5}\times 5 \\ =x^2+\frac{26}{5}x+1 \end{gathered}$$

(xiii) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$.
$$\begin{gathered} \left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right) \\ =z^2+\left(\frac{3}{4}+\frac{4}{3}\right)x+\frac{3}{4}\times \frac{4}{3} \\ =z^2+\frac{25}{12}z+1 \end{gathered}$$

(xiv) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$.
$$\begin{gathered} \left(x^2+4\right)\left(x^2+9\right) \\ ={\left(x^2\right)}^2+\left(4+9\right)\left(x^2\right)+4\times 9 \\ =x^4+13x^2+36 \end{gathered}$$

(xv) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$.
$$\begin{gathered} \left(y^2+12\right)\left(y^2+6\right) \\ ={\left(y^2\right)}^2+\left(12+6\right)\left(y^2\right)+12\times 6 \\ =y^4+18y^2+72 \end{gathered}$$

(xvi) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)=x^2+\left(a-b\right)x-ab$.
$$\begin{gathered} \left(y^2+\frac{5}{7}\right)\left(y^2-\frac{14}{5}\right) \\ ={\left(y^2\right)}^2+\left(\frac{5}{7}-\frac{14}{5}\right)\left(y^2\right)-\frac{5}{7}\times \frac{14}{5} \\ =y^4-\frac{73}{35}{y}^2-2 \end{gathered}$$

(xvii) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)=x^2+\left(a-b\right)x-ab$.
$$\begin{gathered} \left(p^2+16\right)\left(p^2-\frac{1}{4}\right) \\ ={\left(p^2\right)}^2+\left(16-\frac{1}{4}\right)\left(p^2\right)-16\times \frac{1}{4} \\ =p^4+\frac{63}{4}{p}^2-4 \end{gathered}$$

Question 2:

Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006

Answer 2:

(i) Here, we will use the identity $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$
$$\begin{gathered} 102\times 106 \\ =\left(100+2\right)\left(100+6\right) \\ ={100}^2+\left(2+6\right)100+2\times 6 \\ =10000+800+12 \\ =10812 \end{gathered}$$

(ii) Here, we will use the identity $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$
$$\begin{gathered} \mathrm{109\; \times \; 107} \\ =\left(100+9\right)\left(100+7\right) \\ ={100}^2+\left(9+7\right)100+9\times 7 \\ =10000+1600+63 \\ =11663 \end{gathered}$$

(iii) Here, we will use the identity $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$
$$\begin{gathered} \mathrm{35\; \times \; 37} \\ =\left(30+5\right)\left(30+7\right) \\ ={30}^2+\left(5+7\right)30+5\times 7 \\ =900+360+35 \\ =1295 \end{gathered}$$

(iv) Here, we will use the identity $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$
$$\begin{gathered} \mathrm{53\; \times \; 55} \\ =\left(50+3\right)\left(50+5\right) \\ ={50}^2+\left(3+5\right)50+3\times 5 \\ =2500+400+15 \\ =2915 \end{gathered}$$

(v) Here, we will use the identity $\left(x+a\right)\left(x-b\right)=x^2+\left(a-b\right)x-ab$
$$\begin{gathered} \mathrm{103\; \times \; 96} \\ =\left(100+3\right)\left(100-4\right) \\ ={100}^2+\left(3-4\right)100-3\times 4 \\ =10000-100-12 \\ =9888 \end{gathered}$$

(vi) Here, we will use the identity $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$
$$\begin{gathered} \mathrm{34\; \times \; 36} \\ =\left(30+4\right)\left(30+6\right) \\ ={30}^2+\left(4+6\right)30+4\times 6 \\ =900+300+24 \\ =1224 \end{gathered}$$

(vii) Here, we will use the identity $\left(x-a\right)\left(x+b\right)=x^2+\left(b-a\right)x-ab$
$$\begin{gathered} \mathrm{994\; \times \; 1006} \\ =\left(1000-6\right)\times \left(1000+6\right) \\ ={1000}^2+\left(6-6\right)\times 1000-6\times 6 \\ =1000000-36 \\ =999964 \end{gathered}$$