RD Sharma Solutions Class 8 Algebraic Expressions And Identities Exercise 6.5

RD Sharma Solutions Class 8 Chapter 6 Exercise 6.5

RD Sharma Class 8 Solutions Chapter 6 Ex 6.5 PDF Free Download

RD Sharma Solutions Class 8 Chapter 6 Exercise 6.5

Multiply

Q1. (5x + 3) by (7x + 2)

SOLUTION:

To multiply, we will use distributive law as follows:

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

Thus, the answer is \(35 x^{2} + 31 x + 6\)

Q2. (2x + 8) by (x – 3)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 2 x + 8 \right )\left ( x – 3 \right )\\= 2 x \left ( x – 3 \right ) + 8 \left ( x – 3 \right )\\= \left ( 2 x \times x – 2 x \times 3 \right ) + \left ( 8 x – 8 \times 3 \right )\\= \left ( 2 x^{2} – 6 x \right ) + \left ( 8 x – 24 \right )\\= 2 x^{2} – 6 x + 8 x – 24\\= 2 x^{2} + 2 x – 24\)

Thus, the answer is \(2 x^{2} + 2 x – 24\).

Q3. (7x + y) by (x + 5y)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 7 x + y \right )\left ( x + 5 y \right )\\= 7 x \left ( x + 5 y \right )+ y \left ( x + 5 y \right )\\= 7 x^{2} + 35 x y + x y + 5 y^{2}\\= 7 x ^{2} + 36 x y + 5 y^{2}\)

Thus, the answer is \(7 x ^{2} + 36 x y + 5 y^{2}\).

 

 

Q4. (a – 1) by (0.1a2 + 3)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( a – 1 \right )\left ( 0.1 a^{2} + 3 \right )\\= 0.1 a^{2} \left ( a – 1 \right ) + 3 \left ( a – 1 \right )\\= 0.1 a^{3} – 0.1 a^{2} + 3 a – 3\)

Thus, the answer is \(0.1 a^{3} – 0.1 a^{2} + 3 a – 3\).

 

 

Q5. \(\left ( 3 x^{2} + y^{2} \right ) \left ( 2 x^{2} + 3 y^{2} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

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

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

 

 

Q6. \(\left ( \frac{3}{5} x + \frac{1}{2} y \right )\;by\;\left ( \frac{5}{6} x + 4 y \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

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

Thus, the answer is \(\frac{1}{2 } x^{2} + \left ( \frac{169}{60} \right ) x y + 2 y^{2}\).

 

 

Q7. \(\left ( x^{6} – y^{6} \right ) \left ( x^{2} + y^{2} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( x^{6} – y^{6} \right ) \left ( x^{2} + y^{2} \right )\\= x^{6} \left ( x^{2} + y^{2} \right ) – y^{6} \left ( x^{2} + y^{2} \right )\\= \left ( x^{8} + x^{6} y^{2} \right ) – \left ( y^{6} x^{2} + y^{8} \right )\\= x^{8} + x^{6} y^{2} – y^{6} x^{2} – y^{8}\)

Thus, the answer is \(x^{8} + x^{6} y^{2} – y^{6} x^{2} – y^{8}\).

 

 

Q8. \(\left ( x^{2} + y^{2} \right ) \left ( 3 a + 2 b \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

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

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

 

 

Q9. \(\left [ – 3 d + \left ( – 7 f \right ) \right ]\left ( 5 d + f \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left [ – 3 d + \left ( – 7 f \right ) \right ]\left ( 5 d + f \right )\\= \left ( – 3 d \right )\left ( 5 d + f \right ) + \left ( – 7 f \right )\left ( 5 d + f \right )\\= \left ( – 15 d^{2} – 3 d f \right ) + \left ( – 35 d f – 7 f^{2} \right )\\= – 15 d^{2} – 3 d f – 35 d f – 7 f^{2}\\= – 15 d^{2} – 38 d f – 7 f^{2}\)

Thus, the answer is \(- 15 d^{2} – 38 d f – 7 f^{2}\).

 

 

Q10. \(\left ( 0.8 a – 0.5 b \right ) \left ( 1.5 a – 3 b \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 0.8 a – 0.5 b \right ) \left ( 1.5 a – 3 b \right )\\= 0.8 a \left ( 1.5 a – 3 b \right ) – 0.5 b \left ( 1.5 a – 3 b \right )\\= 1.2 a^{2} – 2.4 a b – 0.75 a b + 1.5 b^{2}\\= 1.2 a^{2} – 3.15 a b + 1.5 b^{2}\)

Thus, the answer is \(1.2 a^{2} – 3.15 a b + 1.5 b^{2}\).

 

 

Q11. \(\left ( 2 x^{2} y^{2} – 5 x y^{2} \right ) \left ( x^{2} – y^{2} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 2 x^{2} y^{2} – 5 x y^{2} \right ) \left ( x^{2} – y^{2} \right )\\= 2 x^{2} y^{2} \left ( x^{2} – y^{2} \right ) – 5 x y^{2} \left ( x^{2} – y^{2} \right )\\= 2 x^{4} y^{2} – 2 x^{2} y^{4} – 5 x^{3} y^{2} + 5 x y^{4}\)

Thus, the answer is \(2 x^{4} y^{2} – 2 x^{2} y^{4} – 5 x^{3} y^{2} + 5 x y^{4}\).

 

 

Q12. \(\left ( \frac{x}{7} + \frac{x^{2}}{2} \right ) \left ( \frac{2}{5} + \frac{9x}{4} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( \frac{x}{7} + \frac{x^{2}}{2} \right ) \left ( \frac{2}{5} + \frac{9x}{4} \right )\\= \frac{x}{7} \left ( \frac{2}{5} + \frac{9x}{4} \right ) + \frac{x^{2}}{2} \left ( \frac{2}{5} + \frac{9x}{4} \right )\\= \frac{2x}{35} + \frac{9x^{2}}{28} + \frac{x^{2}}{5} + \frac{9x^{3}}{8}\\= \frac{2x}{35} + \left ( \frac{45 + 28}{140} \right ) x^{2} + \frac{9x^{3}}{8}\\= \frac{2x}{35} + \frac{73}{140} x^{2} + \frac{9x^{3}}{8}\)

Thus, the answer is \(\frac{2x}{35} + \frac{73}{140} x^{2} + \frac{9x^{3}}{8}\)

 

 

Q13. \(\left (- \frac{a}{7} + \frac{a^{2}}{9} \right ) \left ( \frac{b}{2} – \frac{b^{2}}{3} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left (- \frac{a}{7} + \frac{a^{2}}{9} \right ) \left ( \frac{b}{2} – \frac{b^{2}}{3} \right )\\= \left ( – \frac{a}{7} \right ) \left ( \frac{b}{2} – \frac{b^{2}}{3} \right ) + \left ( \frac{a^{2}}{9} \right ) \left ( \frac{b}{2} – \frac{b^{2}}{3} \right )\\= \left ( – \frac{a b}{14} + \frac{a b^{2}}{21} \right ) + \left ( \frac{a^{2} b}{18} – \frac{a^{2} b^{2}}{27} \right )\\= – \frac{a b}{14} + \frac{a b^{2}}{21} + \frac{a^{2} b}{18} – \frac{a^{2} b^{2}}{27}\)

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

 

 

Q14. \(\left ( 3 x^{2} y – 5 x y^{2} \right )\left ( \frac{1}{5} x^{2} + \frac{1}{3} y^{2} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 3 x^{2} y – 5 x y^{2} \right )\left ( \frac{1}{5} x^{2} + \frac{1}{3} y^{2} \right )\\= \frac{1}{5} x^{2} \left ( 3 x^{2} y – 5 x y^{2} \right ) + \frac{1}{3} y^{2} \left ( 3 x^{2} y – 5 x y^{2} \right )\\= \frac{3}{5} x^{4} y – x^{3} y^{2} + x^{2} y^{3} – \frac{5}{3} x y^{4}\)

Thus, the answer is \(\frac{3}{5} x^{4} y – x^{3} y^{2} + x^{2} y^{3} – \frac{5}{3} x y^{4}\).

 

 

Q15. \(\left ( 2 x^{2} – 1 \right ) \left ( 4 x^{3} + 5 x^{2} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 2 x^{2} – 1 \right ) \left ( 4 x^{3} + 5 x^{2} \right )\\= 2 x^{2} \left ( 4 x^{3} + 5 x^{2} \right ) – 1 \left ( 4 x^{3} + 5 x^{2} \right )\\= 8 x^{5} + 10 x^{4} – 4 x^{3} – 5 x^{2}\)

Thus, the answer is \(8 x^{5} + 10 x^{4} – 4 x^{3} – 5 x^{2}\).

 

 

Q16. \(\left ( 2 x y + 3 y^{2} \right ) \left ( 3 y^{2} – 2 \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 2 x y + 3 y^{2} \right ) \left ( 3 y^{2} – 2 \right )\\= 2 x y \left ( 3 y^{2} – 2 \right ) + 3 y^{2} \left ( 3 y^{2} – 2 \right )\\= 6 x y^{3} – 4 x y + 9 y^{4} – 6 y^{2}\\= 9 y^{4} + 6 x y^{3} – 6 y^{2} – 4 x y\)

Thus, the answer is \(9 y^{4} + 6 x y^{3} – 6 y^{2} – 4 x y\).

Find the following products and verify the result for x = -1 and y = -2:

 

 

Q17. \(\left ( 3 x – 5 y \right ) \left ( x + y \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( 3 x – 5 y \right ) \left ( x + y \right )\\= 3 x \left ( x + y \right ) – 5 y \left ( x + y \right )\\= 3 x^{2} + 3 x y – 5 x y – 5 y^{2}\\= 3 x^{2} – 2 x y – 5 y^{2}\\=∴ \left ( 3 x – 5 y \right ) \left ( x + y \right ) = 3 x^{2} – 2 x y – 5 y^{2}\)

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

LHS = \(\left ( 3 x – 5 y \right ) \left ( x + y \right )\\= \left \{ 3 \left ( -1 \right ) – 5 \left ( -2 \right ) \right \}\left \{ -1 + \left ( -2 \right ) \right \}\\= \left ( -3 + 10 \right ) \left ( -3 \right )\\= -21\)

RHS = \(3 x^{2} – 2 x y – 5 y^{2}\\= 3 \left ( -1 \right )^{2} – 2 \left ( -1 \right ) \left ( -2 \right ) – 5 \left ( -2 \right )^{2}\\= 3 \times 1 – 4 – 5 \times 4\\= 3 – 4 – 20\\= -21\)

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

 

 

Q18. \(\left ( x^{2} y – 1 \right ) \left ( 3 – 2 x^{2} y \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( x^{2} y – 1 \right ) \left ( 3 – 2 x^{2} y \right )\\= x^{2} y \left ( 3 – 2 x^{2} y \right ) – 1 \left ( 3 – 2 x^{2} y \right )\\= 3 x^{2} y – 2 x^{4} y^{2} – 3 + 2 ^{2} y\\= 5 x^{2} y – 2 x^{4} y^{2} – 3\\ ∴ \left ( x^{2} y – 1 \right ) \left ( 3 – 2 x^{2} y \right ) = 5 x^{2} y – 2 x^{4} y^{2} – 3\)

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

LHS = \(\left ( x^{2} y – 1 \right ) \left ( 3 – 2 x^{2} y \right )\\= \left [ \left ( -1 \right )^{2} \left ( -2 \right ) – 1 \right ]\left [ 3 – 2 \left ( -1 \right )^{2} \left ( -2 \right ) \right ]\\= \left [ 1 \times \left ( -2 \right ) – 1 \right ]\left [ 3 – 2 \times 1 \times \left ( -2 \right ) \right ]\\= \left ( -2 – 1 \right ) \left ( 3 + 4 \right )\\= -3 \times 7\\= -21\)

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

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

 

 

Q19. \(\left ( \frac{1}{3} x – \frac{y^{2}}{5} \right ) \left ( \frac{1}{3} x + \frac{y^{2}}{5} \right )\)

SOLUTION:

To multiply, we will use distributive law as follows:

\(\left ( \frac{1}{3} x – \frac{y^{2}}{5} \right ) \left ( \frac{1}{3} x + \frac{y^{2}}{5} \right )\\= \left [ \frac{1}{3} x \left ( \frac{1}{3} x + \frac{y^{2}}{5} \right ) \right ] – \left [ \frac{y^{2}}{5} \left ( \frac{1}{3} x + \frac{y^{2}}{5} \right )\right ]\\= \left [ \frac{1}{9} x^{2} + \frac{x y^{2}}{15}\right ] – \left [ \frac{x y^{2}}{15} + \frac{y^{4}}{25} \right ]\\= \frac{1}{9} x^{2} + \frac{x y^{2}}{15} – \frac{x y^{2}}{15} – \frac{y^{4}}{25}\\= \frac{1}{9} x^{2} – \frac{y^{4}}{25}\\ ∴ \left ( \frac{1}{3} x – \frac{y^{2}}{5} \right ) \left ( \frac{1}{3} x + \frac{y^{2}}{5} \right ) = \frac{1}{9} x^{2} – \frac{y^{4}}{25}\)

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

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

RHS = \(\frac{1}{9} x^{2} – \frac{y^{4}}{25}\\= \frac{1}{9} \left ( -1 \right )^{2} – \frac{\left ( -2 \right )^{4}}{25}\\= \frac{1}{9} \times 1 – \frac{16}{25}\\= \frac{1}{9} – \frac{16}{25}\\= – \frac{119}{225}\)

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

 

 

Simplify:

Q20. \(x^{2} \left ( x + 2 y \right ) \left ( x – 3 y \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(x^{2} \left ( x + 2 y \right ) \left ( x – 3 y \right )\\= \left [ x^{2} \left ( x + 2 y \right ) \right ]\left ( x – 3 y \right )\\= \left ( x^{3} + 2 x^{2} y \right ) \left ( x – 3 y \right )\\= x^{3} \left ( x – 3 y \right ) + 2 x^{2} y \left ( x – 3 y \right )\\= x^{4} – 3 x^{3} + 2 x^{3} – 6 x^{2} y^{2}\\= x^{4} – x^{3} – 6 x^{2} y^{2}\)

Thus, the answer is \(x^{4} – x^{3} – 6 x^{2} y^{2}\).

 

 

Q21. \(\left ( x^{2} – 2 y^{2} \right ) \left ( x + 4 y \right ) x^{2} y^{2}\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( x^{2} – 2 y^{2} \right ) \left ( x + 4 y \right ) x^{2} y^{2}\\= \left [ x^{2} \left ( x + 4 y \right ) – 2 y^{2} \left ( x + 4 y \right ) \right ] x^{2} y^{2}\\= \left ( x^{3} + 4 x^{2} y – 2 x y^{2} – 8 y^{3} \right ) x^{2} y^{2}\\= x^{5} y^{2} + 4 x^{4} y^{3} – 2 x^{3} y^{4} – 8 x^{2} y^{5}\)

Thus, the answer is \(x^{5} y^{2} + 4 x^{4} y^{3} – 2 x^{3} y^{4} – 8 x^{2} y^{5}\).

 

 

Q22. \(a^{2} b^{2} \left ( a + 2 b \right ) \left ( 3 a + b \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

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

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

 

 

Q23. \(x^{2} \left ( x – y \right ) y^{2} \left ( x + 2 y \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(x^{2} \left ( x – y \right ) y^{2} \left ( x + 2 y \right )\\= \left [ x^{2} \left ( x – y \right ) \right ] \left [ y^{2} \left ( x + 2 y \right ) \right ]\\= \left ( x^{3} – x^{2} y \right ) \left ( x y^{2} + 2 y^{3} \right )\\= x^{3} \left ( x y^{2} + 2 y^{3} \right ) – x^{2} y \left ( x y^{2} + 2 y^{3} \right )\\= x^{4} y^{2} + 2 x^{3} y^{3} – \left [ x^{3} y^{3} + 2 x^{2} y^{4} \right ]\\= x^{4} y^{2} + 2 x^{3} y^{3} – x^{3} y^{3} – 2 x^{2} y^{4}\\= x^{4} y^{2} + x^{3} y^{3} – 2 x^{2} y^{4}\)

Thus, the answer is \(x^{4} y^{2} + 2 x^{3} y^{3} – x^{3} y^{3} – 2 x^{2} y^{4}\\= x^{4} y^{2} + x^{3} y^{3} – 2 x^{2} y^{4}\)

 

 

Q24. \(\left ( x^{3} – 2 x^{2} + 5 x – 7 \right ) \left ( 2 x – 3 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( x^{3} – 2 x^{2} + 5 x – 7 \right ) \left ( 2 x – 3 \right )\\= 2 x \left ( x^{3} – 2 x^{2} + 5 x – 7 \right ) – 3 \left ( x^{3} – 2 x^{2} + 5 x – 7 \right )\\= 2 x^{4} – 4 x^{3} + 10 x^{2} – 14 x – 3 x^{3} + 6 x^{2} – 15 x + 21\\= 2 x^{4} – 4 x^{3} – 3 x^{3} + 10 x^{2} + 6 x^{2} – 14 x – 15 x + 21\\= 2 x^{4} – 7 x^{3} + 16 x^{2} – 29 x + 21\)

Thus, the answer is \(2 x^{4} – 7 x^{3} + 16 x^{2} – 29 x + 21\).

 

 

Q25. \(2 x^{4} – 7 x^{3} + 16 x^{2} – 29 x + 21\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( 5 x + 3 \right ) \left ( x – 1 \right ) \left ( 3 x – 2 \right )\\= \left [ \left ( 5 x + 3 \right ) \left ( x – 1 \right ) \right ] \left ( 3 x – 2 \right )\\= \left [ 5 x \left ( x – 1 \right ) + 3 \left ( x – 1 \right ) \right ] \left ( 3 x – 2 \right )\\= \left [5 x^{2} – 5 x + 3 x – 3 \right ]\left ( 3 x – 2 \right )\\= 3 x \left ( 5 x^{2} + 2 x – 3 \right ) – 2 \left (5 x^{2} + 2 x – 3 \right )\\= 15 x^{3} – 6 x^{2} – 9 x – \left [ 10 x^{2} – 4 x – 6 \right ]\\= 15 x^{3} – 6 x^{2} – 9 x – 10 x^{2} + 4 x + 6\\= 15 x^{3} – 16 x^{2} – 5 x + 6\)

Thus, the answer is \(15 x^{3} – 6 x^{2} – 9 x – 10 x^{2} + 4 x + 6\\= 15 x^{3} – 16 x^{2} – 5 x + 6\).

 

 

Q26. \(\left ( 5 – x \right )\left ( 6 – 5 x \right )\left ( 2 – x \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( 5 – x \right )\left ( 6 – 5 x \right )\left ( 2 – x \right )\\= \left [ \left ( 5 – x \right )\left ( 6 – 5 x \right ) \right ]\left ( 2 – x \right )\\= \left [ 5 \left ( 6 – 5 x \right ) – x \left ( 6 – 5 x \right ) \right ]\left ( 2 – x \right )\\= \left ( 30 – 25 x – 6 x + 5 x^{2} \right )\left ( 2 – x \right )\\= \left ( 30 – 31 x + 5 x^{2} \right )\left ( 2 – x \right )\\= 2 \left (30 – 31 x + 5 x^{2} \right ) – x \left ( 30 – 31 x + 5 x^{2} \right )\\= 60 – 62 x + 10 x^{2} – 30 x + 31 x^{2} – 5 x^{3}\\= 60 – 92 x + 41 x^{2} – 5 x^{3}\)

Thus, the answer is \(60 – 92 x + 41 x^{2} – 5 x^{3}\).

 

 

Q27. \(\left ( 2 x^{2} + 3 x – 5 \right )\left ( 3 x^{2} – 5 x + 4 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( 2 x^{2} + 3 x – 5 \right )\left ( 3 x^{2} – 5 x + 4 \right )\\= 2 x^{2}\left ( 3 x^{2} – 5 x + 4 \right ) + 3 x \left ( 3 x^{2} – 5 x + 4 \right ) – 5 \left ( 3 x^{2} – 5 x + 4 \right )\\= 6 x^{4} – 10 x^{3} + 8 x^{2} + 9 x^{3} – 15 x^{2} + 12 x – 15 x^{2} + 25 x – 20\\= 6 x^{4} – 10 x^{3} + 9 x^{3} + 8 x^{2} – 15 x^{2} – 15 x^{2} + 25 x + 12 x – 20\\= 6 x^{4} – x^{3} – 22 x^{2} + 36 x – 20\)

Thus, the answer is \(6 x^{4} – x^{3} – 22 x^{2} + 36 x – 20\).

 

 

Q28. \(\left ( 3 x – 2 \right )\left ( 2 x – 3 \right ) + \left ( 5 x – 3 \right )\left ( x + 1 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( 3 x – 2 \right )\left ( 2 x – 3 \right ) + \left ( 5 x – 3 \right )\left ( x + 1 \right )\\= \left [ \left ( 3 x – 2 \right )\left ( 2 x – 3 \right ) \right ] + \left [ \left ( 5 x – 3 \right )\left ( x + 1 \right ) \right ]\\= \left [ 3 x \left ( 2 x – 3 \right ) – 2 \left ( 2 x – 3 \right ) \right ] + \left [ 5 x \left ( x + 1 \right ) – 3 \left ( x + 1 \right ) \right ]\\= 6 x^{2} – 9 x – 4 x + 6 + 5 x^{2} + 5 x – 3 x – 3\\= 6 x^{2} + 5 x^{2} – 9 x – 4 x + 5 x – 3 x – 3 + 6\\= 11 x^{2} – 11 x + 3\)

Thus, the answer is \(11 x^{2} – 11 x + 3\).

 

 

Q29. \(\left ( 5 x – 3 \right )\left (  x + 2 \right ) – \left ( 2 x + 5 \right )\left (4 x – 3 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( 5 x – 3 \right )\left (  x + 2 \right ) – \left ( 2 x + 5 \right )\left ( 4x – 3 \right )\\= \left [ \left ( 5 x – 3 \right )\left ( x + 2 \right ) \right ] – \left [ \left ( 2 x + 5 \right )\left ( 4x – 3 \right ) \right ]\\= \left [ 5 x \left (  x + 2 \right ) – 3 \left ( x + 2 \right ) \right ] – \left [ 2 x \left (4 x – 3 \right ) + 5 \left ( 4x – 3 \right ) \right ]\\= 5 x^{2} + 10 x – 3 x – 6 + 8 x^{2} + 6 x – 20 x + 15\\= 5 x^{2} – 8 x^{2} + 10 x – 3 x + 6 x- 20 x – 6 + 15\\= -3 x^{2} – 7 x + 9\)

Thus, the answer is \(-3 x^{2} – 7 x + 9\).

 

 

Q30. \(\left ( 3 x + 2 y \right ) \left ( 4 x + 3 y \right ) – \left ( 2 x – y \right ) \left ( 7 x – 3 y \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( 3 x + 2 y \right ) \left ( 4 x + 3 y \right ) – \left ( 2 x – y \right ) \left ( 7 x – 3 y \right )\\= \left [ \left ( 3 x + 2 y \right ) \left ( 4 x + 3 y \right ) \right ] – \left [ \left ( 2 x – y \right ) \left ( 7 x – 3 y \right )\right ]\\= \left [ 3 x \left ( 4 x + 3 y \right ) + 2 y \left ( 4 x + 3 y \right ) \right ] – \left [ 2 x \left ( 7 x – 3 y \right ) – y \left ( 7 x – 3 y \right )\right ]\\= 12 x^{2} + 9 x y + 8 x y + 6 y^{2} – 14 x^{2} + 6 x y + 7 x y – 3 y^{2}\\= 12 x^{2} – 14 x^{2} + 9 x y + 8 x y + 6 x y + 7 x y + 6 y^{2} – 3 y^{2}\\= – 2 x^{2} + 30 x y + 3 y^{2}\)

Thus, the answer is \(- 2 x^{2} + 30 x y + 3 y^{2}\).

 

 

Q31. \(\left ( x^{2} – 3 x + 2 \right ) \left ( 5 x – 2 \right ) – \left ( 3 x^{2} + 4 x – 5 \right ) \left ( 2 x – 1 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( x^{2} – 3 x + 2 \right ) \left ( 5 x – 2 \right ) – \left ( 3 x^{2} + 4 x – 5 \right ) \left ( 2 x – 1 \right )\\= \left [ \left ( x^{2} – 3 x + 2 \right ) \left ( 5 x – 2 \right )\right ] – \left [ \left ( 3 x^{2} + 4 x – 5 \right ) \left ( 2 x – 1 \right ) \right ]\\= \left [ 5 x \left ( x^{2} – 3 x + 2 \right ) – 2 \left ( x^{2} – 3 x + 2 \right ) \right ] – \left [ 2 x \left ( 3 x^{2} + 4 x – 5 \right ) – 1 \left ( 3 x^{2} + 4 x – 5 \right )\right ]\\= \left [ 5 x^{3} – 15 x^{2} + 10 x – 2 x^{2} + 6 x – 4 \right ] – \left [ 6 x^{3} + 8 x^{2} – 10 x – 3 x^{2} – 4 x + 5 \right ]\\= 5 x^{3} – 15 x^{2} + 10 x – 2 x^{2} + 6 x – 4 – 6 x^{3} – 8 x^{2} + 10 x + 3 x^{2} + 4 x – 5\\= – x^{3} – 22 x^{2} + 30 x – 9\)

Thus, the answer is \(- x^{3} – 22 x^{2} + 30 x – 9\).

 

 

Q32. \(\left ( x^{3} – 2 x^{2} + 3 x – 4 \right ) \left ( x – 1 \right ) – \left ( 2 x – 3 \right ) \left ( x^{2} – x + 1 \right )\)

SOLUTION:

To simplify, we will use distributive law as follows:

\(\left ( x^{3} – 2 x^{2} + 3 x – 4 \right ) \left ( x – 1 \right ) – \left ( 2 x – 3 \right ) \left ( x^{2} – x + 1 \right )\\= \left [ \left ( x^{3} – 2 x^{2} + 3 x – 4 \right ) \left ( x – 1 \right ) \right ] – \left [ \left ( 2 x – 3 \right ) \left ( x^{2} – x + 1 \right ) \right ]\\= \left [ x \left ( x^{3} – 2 x^{2} + 3 x – 4 \right ) – 1 \left ( x^{3} – 2 x^{2} + 3 x – 4 \right )\right ] – \left [ 2 x \left ( x^{2} – x + 1 \right ) – 3 \left ( x^{2} – x + 1 \right ) \right ]\\= x^{4} – 2 x ^{3} + 3 x^{2} – 4 x – x^{3} + 2 x^{2} – 3 x + 4 – 2 x^{3} + 2 x^{2} – 2 x + 3 x^{2} – 3 x + 3\\= x^{4} – 2 x ^{3} – 2 x^{3} – x^{3} + 3 x^{2} + 2 x^{2} + 2 x^{2} + 3 x^{2} – 4 x – 3 x – 2 x – 3 x + 4 + 3\\= x^{4} – 5 x^{3} + 10 x^{2} – 12 x + 7\)

Thus, the answer is \(x^{4} – 5 x^{3} + 10 x^{2} – 12 x + 7\)..

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