Solve Exercises Lesson 4. Newton binomial (C5 – Math 10 Kite)==============

Solution of Exercises Lesson 1, page 19, Math Textbook 10, Kite episode 2

Expand the following expressions:

a) $${\left( {2x + 1} \right)^4}$$

b)$${\left( {3y – 4} \right)^4}$$

c)$${\left( {x + \frac{1}{2}} \right)^4}$$

d)$${\left( {x – \frac{1}{3}} \right)^4}$$

Solution method

Use Newton’s binomial expansion with $$n = 4$$: $${\left( {a + b} \right)^4} = {a^4} + 4{a^3}b +6{ a^2}{b^2} + 4a{b^3} + {b^4}$$

Solution guide

a) $${\left( {2x + 1} \right)^4} = {\left( {2x} \right)^4} + 4. {\left( {2x} \right)^3}{ .1^1} + 6. {\left( {2x} \right)^2}{.1^2} + 4.\left( {2x} \right){.1^3} + {1^4 } = 16{x^4} + 32{x^3} + 24{x^2} + 8x + 1$$

b) $$\begin{array}{l}{\left( {3y – 4} \right)^4} = {\left[ {3y + \left( { – 4} \right)} \right]^4} = {\left( {3y} \right)^4} + 4. {\left( {3y} \right)^3}.\left( { – 4} \right) + 6. {\left ( {3y} \right)^2}. {\left( { – 4} \right)^2} + 4. {\left( {3y} \right)^1}{\left( { – 4} \right)^3} + {\left( { – 4} \right)^4}\\ = 81{y^4} – 432{y^3} + 864{y^2} – 768y + 256\end{array}$$

c) $${\left( {x + \frac{1}{2}} \right)^4} = {x^4} + 4. {x^3}. {\left( {\frac{1 }{2}} \right)^1} + 6. {x^2}. {\left( {\frac{1}{2}} \right)^2} + 4.x. {\left( { \frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^4} = {x^4} + 2{x^3} + \frac{3}{2}{x^2} + \frac{1}{2}x + \frac{1}{{16}}$$

d) $$\begin{array}{l}{\left( {x – \frac{1}{3}} \right)^4} = {\left[ {x + \left( { – \frac{1}{3}} \right)} \right]^4} = {x^4} + 4. {x^3}. {\left( { – \frac{1}{3}} \right)^1} + 6. {x^2}. {\left( { – \frac{1}{3}} \right)^2} + 4.x. {\left( { – \frac{1}{3}} \right)^3} + {\left( { – \frac{1}{3}} \right)^4}\\ = {x^4 } – \frac{4}{3}{x^3} + \frac{2}{3}{x^2} – \frac{4}{9}x + \frac{1}{{81}} \end{array}$$

Solution of Exercises Lesson 2, page 19, Math textbook 10 Kite, episode 2

Expand the following expressions:

a) $${\left( {x + 1} \right)^5}$$

b) $${\left( {x – 3y} \right)^5}$$

Solution method

Use Newton’s binomial expansion with $$n = 5$$:$${\left( {a + b} \right)^5} = {a^5} + 5{a^4}b + 10{ a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5}$$

Solution guide

a) $${\left( {x + 1} \right)^5} = {x^5} + 5. {x^4}.1 + 10. {x^3}{.1^2} + 10. {x^2}{.1^3} + 5. {x^1}{.1^4} +{1^5} = {x^5} + 5{x^4} + 10{x ^3} + 10{x^2} + 5x + 1$$

b) $$\begin{array}{l}{\left( {x – 3y} \right)^5} = {\left[ {x + \left( { – 3y} \right)} \right]^5} = {x^5} + 5{x^4}{\left( { – 3y} \right)^1} + 10{x^3}{\left( { – 3y} \right)^2 } + 10{x^2}{\left( { – 3y} \right)^3} + 5{x^1}{\left( { – 3y} \right)^4} + {\left( { – 3y} \right)^5}\\ = {x^5} – 15{x^4}y + 90{x^3}{y^2} – 270{x^2}{y^3} + 405x {y^4} – 243{y^5}\end{array}$$

Solve the exercises Lesson 3 page 19 Math textbook 10 Kite episode 2

Determine the coefficient of $${x^4}$$ in the expression expansion $${\left( {3x + 2} \right)^5}$$

Solution method

B1: Use Newton’s binomial expansion with $$n = 5$$:$${\left( {a + b} \right)^5} = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5}$$

B2: Find the coefficient of $${x^4}$$

Solution guide

+) We have: $$\begin{array}{l}{\left( {3x + 2} \right)^5} = {\left( {3x} \right)^5} + 5. {\left ( {3x} \right)^4}2 + 10. {\left( {3x} \right)^3}{2^2} + 10{\left( {3x} \right)^2}{.2 ^3} + 5.\left( {3x} \right){.2^4} + {2^5}\\ = 243{x^5} + 810{x^4} + 1080{x^3} + 720{x^2} + 240x + 32\end{array}$$

+) The coefficient of $${x^4}$$ in the above expansion is: $${a_4} = 810$$

Solve the exercises Lesson 4, page 19, Math textbook 10 Kite, episode 2

Let $${\left( {1 – \frac{1}{2}x} \right)^5} = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x ^3} + {a_4}{x^4} + {a_5}{x^5}$$ . Calculate:

a) $${a_3}$$

b) $${a_0} + {a_1} + {a_2} + {a_3} + {a_4} + {a_5}$$

Solution method

a) Step 1: Use Newton’s binomial expansion with $$n = 5$$:

$${\left( {a + b} \right)^5} = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2} {b^3} + 5a{b^4} + {b^5}$$

Step 2: Identify the coefficient $$\Rightarrow {a_3}$$ as the coefficient of $${x_3}$$

b) Comment: Substituting $$x = 1$$ into the initial expansion we have the sum to be calculated immediately

Solution guide

a) +) We have: $${\left( {1 – \frac{1}{2}x} \right)^5} = 1 – \frac{5}{2}x + \frac{5} {2}{x^2} – \frac{5}{4}{x^3} + \frac{5}{{16}}{x^4} – \frac{1}{{32}}{ x^5}$$

+) Identifying the coefficient with the expansion in the problem, we see: $${a_3} = \frac{{ – 5}}{4}$$

b) +) Replace $$x = 1$$ into the expanded expression in the problem, we have: $${\left( {1 – \frac{1}{2}.1} \right)^5} = {a_0} + {a_1} + {a_2} + {a_3} + {a_4} + {a_5}$$

+) So the sum :$${a_0} + {a_1} + {a_2} + {a_3} + {a_4} + {a_5} = {\left( {\frac{1}{2}} \right)^5 } = \frac{1}{{32}}$$

Solve the exercises Exercise 5 page 19 Math textbook 10 Kite episode 2

Given a set A with 5 elements. What is the number of subsets of A?

Solution method

If a set has $$n$$ elements $$\left( {n \in \mathbb{N}} \right)$$ then the number of subsets of that set is: $${2^n}$$ (subset)

Solution guide

Since set A has 5 elements, the number of subsets of set A is: $${2^5} = 32$$ (subset)