RD Sharma solution class 8 chapter 6 Algebraic Expressions and Identity Exercise 6.5

Exercise 6.5

Page-6.30

Question 1:

Multiply:
(5x + 3) by (7x + 2)

Answer 1:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(5x+3\right)\left(7x+2\right) \\ =5x\left(7x+2\right)+3\left(7x+2\right) \\ =\left(5x\times 7x+5x\times 2\right)+\left(3\times 7x+3\times 2\right) \\ =\left(35x^2+10x\right)+\left(21x+6\right) \\ =35x^2+10x+21x+6 \\ =35x^2+31x+6 \end{gathered}$$

Thus, the answer is $35x^2+31x+6$.

Question 2:

Multiply:
(2x + 8) by (x − 3)

Answer 2:

To multiply the expressions, we will use the distributive law in the following way:

$$\begin{gathered} \left(2x+8\right)\left(x-3\right) \\ =2x\left(x-3\right)+8\left(x-3\right) \\ =\left(2x\times x-2x\times 3\right)+\left(8x-8\times 3\right) \\ =\left(2x^2-6x\right)+\left(8x-24\right) \\ =2x^2-6x+8x-24 \\ =2x^2+2x-24 \end{gathered}$$

Thus, the answer is $2x^2+2x-24$.

Question 3:

Multiply:
(7x + y) by (x + 5y)

Answer 3:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(7x+y\right)\left(x+5y\right) \\ =7x\left(x+5y\right)+y\left(x+5y\right) \\ =7x^2+35xy+xy+5y^2 \\ =7x^2+36xy+5y^2 \end{gathered}$$

Thus, the answer is $7x^2+36xy+5y^2$.

Question 4:

Multiply:
(a − 1) by (0.1a² + 3)

Answer 4:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(a-1\right)\left(0.1a^2+3\right) \\ =0.1a^2\left(a-1\right)+3\left(a-1\right) \\ =0.1a^3-0.1a^2+3a-3 \end{gathered}$$

Thus, the answer is $0.1a^3-0.1a^2+3a-3$.

Question 5:

Multiply:
(3x² + y²) by (2x² + 3y²)

Answer 5:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(3x^2+y^2\right)\left(2x^2+3y^2\right) \\ =3x^2\left(2x^2+3y^2\right)+y^2\left(2x^2+3y^2\right) \\ =6x^4+9x^2{y}^2+2x^2{y}^2+3y^4 \\ =6x^4+11x^2{y}^2+3y^4 \end{gathered}$$

Thus, the answer is $6x^4+11x^2{y}^2+3y^4$.

Question 6:

Multiply:
$\left(\frac{3}{5}x+\frac{1}{2}y\right) \mathrm{by} \left(\frac{5}{6}x+4y\right)$

Answer 6:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(\frac{3}{5}x+\frac{1}{2}y\right)\left(\frac{5}{6}x+4y\right) \\ =\frac{3}{5}x\left(\frac{5}{6}x+4y\right)+\frac{1}{2}y\left(\frac{5}{6}x+4y\right) \\ =\frac{1}{2}{x}^2+\frac{12}{5}xy+\frac{5}{12}xy+2y^2 \\ =\frac{1}{2}{x}^2+\left(\frac{144+25}{60}\right)xy+2y^2 \\ =\frac{1}{2}{x}^2+\frac{169}{60}xy+2y^2 \end{gathered}$$

Thus, the answer is $\frac{1}{2}{x}^2+\frac{169}{60}xy+2y^2$.

Page-6.31

Question 7:

Multiply:
(x⁶ − y⁶) by (x² + y²)

Answer 7:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(x^6-y^6\right)\left(x^2+y^2\right) \\ =x^6\left(x^2+y^2\right)-y^6\left(x^2+y^2\right) \\ =\left(x^8+x^6{y}^2\right)-\left(y^6{x}^2+y^8\right) \\ =x^8+x^6{y}^2-y^6{x}^2-y^8 \end{gathered}$$

Thus, the answer is $x^8+x^6{y}^2-y^6{x}^2-y^8$.

Question 8:

Multiply:
(x² + y²) by (3a + 2b)

Answer 8:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(x^2+y^2\right)\left(3a+2b\right) \\ =x^2\left(3a+2b\right)+y^2\left(3a+2b\right) \\ =3a{x}^2+2b{x}^2+3a{y}^2+2b{y}^2 \end{gathered}$$

Thus, the answer is $3a{x}^2+2b{x}^2+3a{y}^2+2b{y}^2$.

Question 9:

Multiply:
[−3d + (−7f)] by (5d + f)

Answer 9:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left[-3d+\left(-7f\right)\right]\left(5d+f\right) \\ =\left(-3d\right)\left(5d+f\right)+\left(-7f\right)\left(5d+f\right) \\ =\left(-15d^2-3df\right)+\left(-35df-7f^2\right) \\ =-15d^2-3df-35df-7f^2 \\ =-15d^2-38df-7f^2 \end{gathered}$$

Thus, the answer is $-15d^2-38df-7f^2$.

Question 10:

Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

Answer 10:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(0.8a-0.5b\right)\left(1.5a-3b\right) \\ =0.8a\left(1.5a-3b\right)-0.5b\left(1.5a-3b\right) \\ =1.2a^2-2.4ab-0.75ab+1.5b^2 \\ =1.2a^2-3.15ab+1.5b^2 \end{gathered}$$

Thus, the answer is $1.2a^2-3.15ab+1.5b^2$.

Question 11:

Multiply:
(2x²y² − 5xy²) by (x² − y²)

Answer 11:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(2x^2{y}^2-5x{y}^2\right)\left(x^2-y^2\right) \\ =2x^2{y}^2\left(x^2-y^2\right)-5x{y}^2\left(x^2-y^2\right) \\ =2x^4{y}^2-2x^2{y}^4-5x^3{y}^2+5x{y}^4 \end{gathered}$$

Thus, the answer is $2x^4{y}^2-2x^2{y}^4-5x^3{y}^2+5x{y}^4$.

Question 12:

Multiply:
$\left(\frac{x}{7}+\frac{x^2}{2}\right)by\left(\frac{2}{5}+\frac{9x}{4}\right)$

Answer 12:

To multiply the expressions, we will use the distributive law in the following way:

$$\begin{gathered} \left(\frac{x}{7}+\frac{x^2}{2}\right)\left(\frac{2}{5}+\frac{9x}{4}\right) \\ =\frac{x}{7}\left(\frac{2}{5}+\frac{9x}{4}\right)+\frac{x^2}{2}\left(\frac{2}{5}+\frac{9x}{4}\right) \\ =\frac{2x}{35}+\frac{9x^2}{28}+\frac{x^2}{5}+\frac{9x^3}{8} \\ =\frac{2x}{35}+\left(\frac{45+28}{140}\right)x^2+ \frac{9x^3}{8} \\ =\frac{2x}{35}+\frac{73x^2}{140} +\frac{9x^2}{8} \end{gathered}$$
Thus, the answer is  $\frac{2x}{35}+\frac{73x^2}{140}+\frac{9x^3}{8}$

Question 13:

Multiply:
$\left(-\frac{a}{7}+\frac{a^2}{9}\right)by\left(\frac{b}{2}-\frac{b^2}{3}\right)$

Answer 13:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(-\frac{a}{7}+\frac{a^2}{9}\right)\left(\frac{b}{2}-\frac{b^2}{3}\right) \\ =\left(-\frac{a}{7}\right)\left(\frac{b}{2}-\frac{b^2}{3}\right)+\left(\frac{a^2}{9}\right)\left(\frac{b}{2}-\frac{b^2}{3}\right) \\ =\left(-\frac{ab}{14}+\frac{a{b}^2}{21}\right)+\left(\frac{a^{2}b}{18}-\frac{a^2{b}^2}{27}\right) \\ =-\frac{ab}{14}+\frac{a{b}^2}{21}+\frac{a^{2}b}{18}-\frac{a^2{b}^2}{27} \end{gathered}$$

Thus, the answer is $-\frac{ab}{14}+\frac{a{b}^2}{21}+\frac{a^{2}b}{18}-\frac{a^2{b}^2}{27}$.

Question 14:

Multiply:
(3x²y − 5xy²) by $\left(\frac{1}{5}{x}^2 + \frac{1}{3}{y}^2\right)$

Answer 14:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(3x^{2}y-5x{y}^2\right)\left(\frac{1}{5}{x}^2+\frac{1}{3}{y}^2\right) \\ =\frac{1}{5}{x}^2\left(3x^{2}y-5x{y}^2\right)+\frac{1}{3}{y}^2\left(3x^{2}y-5x{y}^2\right) \\ =\frac{3}{5}{x}^{4}y-x^3{y}^2+x^2{y}^3-\frac{5}{3}x{y}^4 \end{gathered}$$

Thus, the answer is $\frac{3}{5}{x}^{4}y-x^3{y}^2+x^2{y}^3-\frac{5}{3}x{y}^4$.

Question 15:

Multiply:
(2x² − 1) by (4x³ + 5x²)

Answer 15:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(2x^2-1\right)\left(4x^3+5x^2\right) \\ =2x^2\left(4x^3+5x^2\right)-1\left(4x^3+5x^2\right) \\ =8x^5+10x^4-4x^3-5x^2 \end{gathered}$$

Thus, the answer is $8x^5+10x^4-4x^3-5x^2$.

Question 16:

(2xy + 3y²) (3y² − 2)

Answer 16:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(2xy+3y^2\right)\left(3y^2-2\right) \\ =2xy\left(3y^2-2\right)+3y^2\left(3y^2-2\right) \\ =6x{y}^3-4xy+9y^4-6y^2 \\ =9y^4+6x{y}^3-6y^2-4xy \end{gathered}$$

Thus, the answer is $9y^4+6x{y}^3-6y^2-4xy$.

Question 17:

Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

Answer 17:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(3x-5y\right)\left(x+y\right) \\ =3x\left(x+y\right)-5y\left(x+y\right) \\ =3x^2+3xy-5xy-5y^2 \\ =3x^2-2xy-5y^2 \end{gathered}$$

$\therefore$ $\left(3x-5y\right)\left(x+y\right)=3x^2-2xy-5y^2$.
Now, we put x = $-$1 and y = $-$2 on both sides to verify the result.

$$\begin{gathered} \text{LHS}=\left(3x-5y\right)\left(x+y\right) \\ =\left\{3\left(-1\right)-5\left(-2\right)\right\}\left\{-1+\left(-2\right)\right\} \\ =\left(-3+10\right)\left(-3\right) \\ =\left(7\right)\left(-3\right) \\ =-21 \end{gathered}$$

$$\begin{gathered} \text{RHS}=3x^2-2xy-5y^2 \\ =3{\left(-1\right)}^2-2\left(-1\right)\left(-2\right)-5{\left(-2\right)}^2 \\ =3\times 1-4-5\times 4 \\ =3-4-20 \\ =-21 \end{gathered}$$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $3x^2-2xy-5y^2$.

Question 18:

Find the following product and verify the result for x = − 1, y = − 2:
(x²y − 1) (3 − 2x²y)

Answer 18:

To multiply, we will  use distributive law as follows:

$$\begin{gathered} \left(x^{2}y-1\right)\left(3-2x^{2}y\right) \\ =x^{2}y\left(3-2x^{2}y\right)-1\times \left(3-2x^{2}y\right) \\ =3x^{2}y-2x^4{y}^2-3+2x^{2}y \\ =5x^{2}y-2x^4{y}^2-3 \end{gathered}$$

$\therefore$ $\left(x^{2}y-1\right)\left(3-2x^{2}y\right)=5x^{2}y-2x^4{y}^2-3$

Now, we put x = $-$1 and y = $-$2 on both sides to verify the result.

$$\begin{gathered} \text{LHS = }\left(x^{2}y-1\right)\left(3-2x^{2}y\right) \\ =\left[{\left(-1\right)}^2\left(-2\right)-1\right]\left[3-2{\left(-1\right)}^2\left(-2\right)\right] \\ =\left[1\times \left(-2\right)-1\right]\left[3-2\times 1\times \left(-2\right)\right] \\ =\left(-2-1\right)\left(3+4\right) \\ =-3\times 7 \\ =-21 \end{gathered}$$

$$\begin{gathered} \text{RHS}=5x^{2}y-2x^4{y}^2-3 \\ =5{\left(-1\right)}^2\left(-2\right)-2{\left(-1\right)}^4{\left(-2\right)}^2-3 \\ =\left[5\times 1\times \left(-2\right)\right]-\left[2\times 1\times 4\right]-3 \\ =-10-8-3 \\ =-21 \end{gathered}$$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $5x^{2}y-2x^4{y}^2-3$.

Question 19:

Find the following product and verify the result for x = − 1, y = − 2:
$\left(\frac{1}{3}x-\frac{y^2}{5}\right)\left(\frac{1}{3}x+\frac{y^2}{5}\right)$

Answer 19:

To multiply, we will use distributive law as follows:

$$\begin{gathered} \left(\frac{1}{3}x-\frac{y^2}{5}\right)\left(\frac{1}{3}x+\frac{y^2}{5}\right) \\ =\left[\frac{1}{3}x\left(\frac{1}{3}x+\frac{y^2}{5}\right)\right]-\left[\frac{y^2}{5}\left(\frac{1}{3}x+\frac{y^2}{5}\right)\right] \\ =\left[\frac{1}{9}{x}^2+\frac{x{y}^2}{15}\right]-\left[\frac{x{y}^2}{15}+\frac{y^4}{25}\right] \\ =\frac{1}{9}{x}^2+\frac{x{y}^2}{15}-\frac{x{y}^2}{15}-\frac{y^4}{25} \\ =\frac{1}{9}{x}^2-\frac{y^4}{25} \end{gathered}$$

$\therefore$ $\left(\frac{1}{3}x-\frac{y^2}{5}\right)\left(\frac{1}{3}x+\frac{y^2}{5}\right)=\frac{1}{9}{x}^2-\frac{y^4}{25}$

Now, we will put x = $-$1 and y = $-$2 on both the sides to verify the result.

$$\begin{gathered} \text{LHS = }\left(\frac{1}{3}x-\frac{y^2}{5}\right)\left(\frac{1}{3}x+\frac{y^2}{5}\right) \\ =\left[\frac{1}{3}\left(-1\right)-\frac{{\left(-2\right)}^2}{5}\right]\left[\frac{1}{3}\left(-1\right)+\frac{{\left(-2\right)}^2}{5}\right] \\ =\left(-\frac{1}{3}-\frac{4}{5}\right)\left(-\frac{1}{3}+\frac{4}{5}\right) \\ =\left(\frac{-17}{15}\right)\left(\frac{7}{15}\right) \\ =\frac{-119}{225} \end{gathered}$$

$$\begin{gathered} \text{RHS}=\frac{1}{9}{x}^2-\frac{y^4}{25} \\ =\frac{1}{9}{\left(-1\right)}^2-\frac{{\left(-2\right)}^4}{25} \\ =\frac{1}{9}\times 1-\frac{16}{25} \\ =\frac{1}{9}-\frac{16}{25} \\ =-\frac{119}{225} \end{gathered}$$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $\frac{1}{9}{x}^2-\frac{y^4}{25}$.

Question 20:

Simplify:
x²(x + 2y) (x − 3y)

Answer 20:

To simplify, we will proceed as follows:

$$\begin{gathered} x^2\left(x+2y\right)\left(x-3y\right) \\ =\left[x^2\left(x+2y\right)\right]\left(x-3y\right) \\ =\left(x^3+2x^{2}y\right)\left(x-3y\right) \\ =x^3\left(x-3y\right)+2x^{2}y\left(x-3y\right) \\ =x^4-3x^{3}y+2x^{3}y-6x^2{y}^2 \\ =x^4-x^{3}y-6x^2{y}^2 \end{gathered}$$

Thus, the answer is $x^4-x^{3}y-6x^2{y}^2$.

Question 21:

Simplify:
(x² − 2y²) (x + 4y) x²y²

Answer 21:

To simplify, we will proceed as follows:

$$\begin{gathered} \left(x^2-2y^2\right)\left(x+4y\right)x^2{y}^2 \\ =\left[x^2\left(x+4y\right)-2y^2\left(x+4y\right)\right]x^2{y}^2 \\ =\left(x^3+4x^{2}y-2x{y}^2-8y^3\right)x^2{y}^2 \\ =x^5{y}^2+4x^4{y}^3-2x^3{y}^4-8x^2{y}^5 \end{gathered}$$

Thus, the answer is $x^5{y}^2+4x^4{y}^3-2x^3{y}^4-8x^2{y}^5$.

Question 22:

Simplify:
a²b²(a + 2b)(3a + b)

Answer 22:

To simplify, we will proceed as follows:

$$\begin{gathered} a^2{b}^2\left(a+2b\right)\left(3a+b\right) \\ =\left[a^2{b}^2\left(a+2b\right)\right]\left(3a+b\right) \\ =\left(a^3{b}^2+2a^2{b}^3\right)\left(3a+b\right) \\ =3a\left(a^3{b}^2+2a^2{b}^3\right)+b\left(a^3{b}^2+2a^2{b}^3\right) \\ =3a^4{b}^2+6a^3{b}^3+a^3{b}^3+2a^2{b}^4 \\ =3a^4{b}^2+7a^3{b}^3+2a^2{b}^4 \end{gathered}$$

Thus, the answer is $3a^4{b}^2+7a^3{b}^3+2a^2{b}^4$.

Question 23:

Simplify:
x²(x − y) y²(x + 2y)

Answer 23:

To simplify, we will proceed as follows:

$$\begin{gathered} x^2\left(x-y\right)y^2\left(x+2y\right) \\ =\left[x^2\left(x-y\right)\right]\left[y^2\left(x+2y\right)\right] \\ =\left(x^3-x^{2}y\right)\left(x{y}^2+2y^3\right) \\ =x^3\left(x{y}^2+2y^3\right)-x^{2}y\left(x{y}^2+2y^3\right) \\ =x^4{y}^2+2x^3{y}^3-\left[x^3{y}^3+2x^2{y}^4\right] \\ =x^4{y}^2+2x^3{y}^3-x^3{y}^3-2x^2{y}^4 \\ =x^4{y}^2+x^3{y}^3-2x^2{y}^4 \end{gathered}$$

Thus, the answer is $x^4{y}^2+x^3{y}^3-2x^2{y}^4$.

Question 24:

Simplify:
(x³ − 2x² + 5x − 7)(2x − 3)

Answer 24:

To simplify, we will proceed as follows:

$$\begin{gathered} \left(x^3-2x^2+5x-7\right)\left(2x-3\right) \\ =2x\left(x^3-2x^2+5x-7\right)-3\left(x^3-2x^2+5x-7\right) \\ =2x^4-4x^3+10x^2-14x-3x^3+6x^2-15x+21 \end{gathered}$$
$=2x^4-4x^3-3x^3+10x^2+6x^2-14x-15x+21$     (Rearranging)
$=2x^4-7x^3+16x^2-29x+21$                              (Combining like terms)

Thus, the answer is $2x^4-7x^3+16x^2-29x+21$.

Question 25:

Simplify:
(5x + 3)(x − 1)(3x − 2)

Answer 25:

To simplify, we will proceed as follows:

$$\begin{gathered} \left(5x+3\right)\left(x-1\right)\left(3x-2\right) \\ =\left[\left(5x+3\right)\left(x-1\right)\right]\left(3x-2\right) \end{gathered}$$
$=\left[5x\left(x-1\right)+3\left(x-1\right)\right]\left(3x-2\right)$              (Distributive law)
$$\begin{gathered} =\left[5x^2-5x+3x-3\right]\left(3x-2\right) \\ =\left[5x^2-2x-3\right]\left(3x-2\right) \\ =3x\left(5x^2-2x-3\right)-2\left(5x^2-2x-3\right) \\ =15x^3-6x^2-9x-\left[10x^2-4x-6\right] \\ =15x^3-6x^2-9x-10x^2+4x+6 \end{gathered}$$
$=15x^3-6x^2-10x^2-9x+4x+6$              (Rearranging)
$=15x^3-16x^2-5x+6$                              (Combining like terms)

Thus, the answer is $15x^3-16x^2-5x+6$.

Question 26:

Simplify:
(5 − x)(6 − 5x)( 2 − x)

Answer 26:

To simplify, we will proceed as follows:

$$\begin{gathered} \left(5-x\right)\left(6-5x\right)\left(2-x\right) \\ =\left[\left(5-x\right)\left(6-5x\right)\right]\left(2-x\right) \end{gathered}$$
$=\left[5\left(6-5x\right)-x\left(6-5x\right)\right]\left(2-x\right)$                (Distributive law)
$$\begin{gathered} =\left(30-25x-6x+5x^2\right)\left(2-x\right) \\ =\left(30-31x+5x^2\right)\left(2-x\right) \\ =2\left(30-31x+5x^2\right)-x\left(30-31x+5x^2\right) \\ =60-62x+10x^2-30x+31x^2-5x^3 \end{gathered}$$
$=60-62x-30x+10x^2+31x^2-5x^3$              (Rearranging)
$=60-92x+41x^2-5x^3$                                (Combining like terms)

Thus, the answer is $60-92x+41x^2-5x^3$.

Question 27:

Simplify:
(2x² + 3x − 5)(3x² − 5x + 4)

Answer 27:

To simplify, we will proceed as follows:

$\left(2x^2+3x-5\right)\left(3x^2-5x+4\right)$
$=2x^2\left(3x^2-5x+4\right)+3x\left(3x^2-5x+4\right)-5\left(3x^2-5x+4\right)$           (Distributive law)
$=6x^4-10x^3+8x^2+9x^3-15x^2+12x-15x^2+25x-20$
$=6x^4-10x^3+9x^3+8x^2-15x^2-15x^2+12x+25x-20$              (Rearranging)
$=6x^4-x^3-22x^2+36x-20$                                                     (Combining like terms)

Thus, the answer is $6x^4-x^3-22x^2+36x-20$.

Question 28:

Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)

Answer 28:

To simplify, we will proceed as follows:

$$\begin{gathered} \left(3x-2\right)\left(2x-3\right)+\left(5x-3\right)\left(x+1\right) \\ =\left[\left(3x-2\right)\left(2x-3\right)\right]+\left[\left(5x-3\right)\left(x+1\right)\right] \end{gathered}$$
$=\left[3x\left(2x-3\right)-2\left(2x-3\right)\right]+\left[5x\left(x+1\right)-3\left(x+1\right)\right]$           (Distributive law)
$=6x^2-9x-4x+6+5x^2+5x-3x-3$
$=6x^2+5x^2-9x-4x+5x-3x-3+6$                                  (Rearranging)
$=11x^2-11x+3$                                                                 (Combining like terms)

Thus, the answer is $11x^2-11x+3$.

Question 29:

Simplify:
(5x − 3)(x + 2) − (2x + 5)(4x − 3)

Answer 29:

To simplify, we will proceed as follows:

$$\begin{gathered} \left(5x-3\right)\left(x+2\right)-\left(2x+5\right)\left(4x-3\right) \\ =\left[\left(5x-3\right)\left(x+2\right)\right]-\left[\left(2x+5\right)\left(4x-3\right)\right] \end{gathered}$$
$=\left[5x\left(x+2\right)-3\left(x+2\right)\right]-\left[2x\left(4x-3\right)+5\left(4x-3\right)\right]$            (Distributive law)
$=5x^2+10x-3x-6-8x^2+6x-20x+15$
$=5x^2-8x^2+10x-3x+6x-20x-6+15$                              (Rearranging)
$$\begin{gathered} =5x^2-8x^2+10x-3x+6x-20x-6+15 \\ =-3x^2-7x+9 \end{gathered}$$                              (Combining like terms)

Hence, the answer is $-3x^2-7x+9$.

Question 30:

Simplify:
(3x + 2y)(4x + 3y) − (2x − y)(7x − 3y)

Answer 30:

To simplify, we will proceed as follows:

$$\begin{gathered} \left(3x+2y\right)\left(4x+3y\right)-\left(2x-y\right)\left(7x-3y\right) \\ =\left[\left(3x+2y\right)\left(4x+3y\right)\right]-\left[\left(2x-y\right)\left(7x-3y\right)\right] \end{gathered}$$
$=\left[3x\left(4x+3y\right)+2y\left(4x+3y\right)\right]-\left[2x\left(7x-3y\right)-y\left(7x-3y\right)\right]$            (Distributive law)
$$\begin{gathered} =12x^2+9xy+8xy+6y^2-\left[14x^2-6xy-7xy+3y^2\right] \\ =12x^2+9xy+8xy+6y^2-14x^2+6xy+7xy-3y^2 \end{gathered}$$
$=12x^2-14x^2+9xy+8xy+6xy+7xy+6y^2-3y^2$                              (Rearranging)
$=-2x^2+30xy+3y^2$                                                                       (Combining like terms)

Thus, the answer is $-2x^2+30xy+3y^2$.

Question 31:

Simplify:
(x² − 3x + 2)(5x − 2) − (3x² + 4x − 5)(2x − 1)

Answer 31:

To simplify, we will  proceed as follows:

$$\begin{gathered} \left(x^2-3x+2\right)\left(5x-2\right)-\left(3x^2+4x-5\right)\left(2x-1\right) \\ =\left[\left(x^2-3x+2\right)\left(5x-2\right)\right]-\left[\left(3x^2+4x-5\right)\left(2x-1\right)\right] \end{gathered}$$
$=\left[5x\left(x^2-3x+2\right)-2\left(x^2-3x+2\right)\right]-\left[2x\left(3x^2+4x-5\right)-1\times \left(3x^2+4x-5\right)\right]$            (Distributive law)
$$\begin{gathered} =\left[5x^3-15x^2+10x-\left(2x^2-6x+4\right)\right]-\left[6x^3+8x^2-10x-3x^2-4x+5\right] \\ =\left[5x^3-15x^2+10x-2x^2+6x-4\right]-\left[6x^3+8x^2-10x-3x^2-4x+5\right] \\ =5x^3-15x^2+10x-2x^2+6x-4-6x^3-8x^2+10x+3x^2+4x-5 \end{gathered}$$
$=5x^3-6x^3-15x^2-2x^2-8x^2+3x^2+10x+6x+10x+4x-5-4$                                (Rearranging)
$=-x^3-22x^2+30x-9$                                                                                            (Combining like terms)

Thus, the answer is $-x^3-22x^2+30x-9$.
 

Question 32:

Simplify:
(x³ − 2x² + 3x − 4) (x −1) − (2x − 3)(x² − x + 1)

Answer 32:

To simplify,we will proceed as follows:

$$\begin{gathered} \left(x^3-2x^2+3x-4\right)\left(x-1\right)-\left(2x-3\right)\left(x^2-x+1\right) \\ =\left[\left(x^3-2x^2+3x-4\right)\left(x-1\right)\right]-\left[\left(2x-3\right)\left(x^2-x+1\right)\right] \end{gathered}$$
$=\left[x\left(x^3-2x^2+3x-4\right)-1\left(x^3-2x^2+3x-4\right)\right]-\left[2x\left(x^2-x+1\right)-3\left(x^2-x+1\right)\right]$            (Distributive law)
$$\begin{gathered} =\left[x\left(x^3-2x^2+3x-4\right)-1\left(x^3-2x^2+3x-4\right)\right]-\left[2x\left(x^2-x+1\right)-3\left(x^2-x+1\right)\right] \\ =x^4-2x^3+3x^2-4x-x^3+2x^2-3x+4-\left[2x^3-2x^2+2x-3x^2+3x-3\right] \\ =x^4-2x^3+3x^2-4x-x^3+2x^2-3x+4-2x^3+2x^2-2x+3x^2-3x+3 \end{gathered}$$
$=x^4-2x^3-2x^3-x^3+3x^2+2x^2+2x^2+3x^2-4x-3x-2x-3x+4+3$                             (Rearranging)
$=x^4-5x^3+10x^2-12x+7$                                                                                            (Combining like terms)

Thus, the answer is $x^4-5x^3+10x^2-12x+7$.