## Solve Exercises Lesson 25: Newton’s binomial (Math 10 – Connect) – Math Book

Solution for Exercise 25: Newton’s binomial (Math 10 – Connection textbook)
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### Solve problem 8.12 page 74 Math textbook 10 Connecting knowledge volume 2

Expanding polynomials:

a) (x -3)4

b) (3x – 2y)4

c) (x+5)4 + (x – 5)4

d) (x – 2y)5

Solution method

Apply the formulas:

$$\begin{array}{l} {\left( {a + b} \right)^4} = {C_4}^0{a^4} + {C_4}^1{a^3}b + {C_4}^2{a^2}{ b^2} + {C_4}^3a{b^3} + {C_4}^4{b^4}\\ \;\;\;\;\;\;\;\;\;\;\;\;\; = {a^4} + 4{a^3}b + 6{a^2}{b^2} + 4a{b^3} + {b^4}. \end{array}$$

$$\begin{array}{*{20}{l}} {{{\left( {a + b} \right)}^5} = {C_4}^0{a^5} + {C_5}^1{a^4}b + {C_5}^2{a^ 3}{b^2} + {C_5}^3{a^2}{b^3} + {C_5}^4a{b^4} + {C_5}^5{b^5}}\\ {\;\;\;\;\;\;\;\;\;\;\;\;\; = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5 }.} \end{array}$$

Detailed explanation

a) (x -3)4 = x4 + 4.x3.(-3) +6.x2.(-3)2 +4.x.(-3)3 + (-3)4

= x4 -12.x3 +54.x2 – 108.x +81.

b) (3x – 2y)4 = (3x)4 + 4.(3x)3(2y) + 6.(3x)2.(2y)2 + 4.(3x).(2y)3 + (2y)4

= 81x4 + 216x3y + 216x2y2 + 96xy3 + 16y4

c) (x+5)4 + (x – 5)4 = (x5+ 5x4.5 + 10x3.52 + 10x2.53 +5x.54+ 55) + (x5+ 5x4.(-5) + 10x3.(-5)2 + 10x2.(-5)3 + 5x.(-5)4+ (-5)5)

= 2x5 + 500x3 + 6250x

d) (x – 2y)5 = x5 +5x4(2y) + 10x3(2y)2 + 10x2(2y)3 +5x(2y)4 + (2y)5

= x5 +10x4y + 40x3y2 + 80x2y3 + 80xy4 + 32y5.

### Solution 8.13 page 74 Math textbook 10 Connecting knowledge volume 2

Find the coefficient of $${x^4}$$ in the expansion of $${\left( {3x – 1} \right)^5}$$

Solution method

Find the term containing x4

Detailed explanation

Term containing x4 is: 5.(3x)4(-1) = -405x4.

So the coefficient of x4 in expansion is: -405.

### Solve problem 8.14 page 74 Math textbook 10 Connecting knowledge volume 2

Represent $$(3+\sqrt{2})^{5}-(3-\sqrt{2})^{5}$$ as $$a+b\sqrt{2}$$ with a, b are integers.

Solution method

Applying formula:

$$\begin{array}{*{20}{l}} {{{\left( {a + b} \right)}^5} = {C_4}^0{a^5} + {C_5}^1{a^4}b + {C_5}^2{a^ 3}{b^2} + {C_5}^3{a^2}{b^3} + {C_5}^4a{b^4} + {C_5}^5{b^5}}\\ {\;\;\;\;\;\;\;\;\;\;\;\;\; = {a^5} + 5{a^4}b + 10{a^3}{b^2} + 10{a^2}{b^3} + 5a{b^4} + {b^5 }.} \end{array}$$

Detailed explanation

$$\begin{array}{l} {(3 + \sqrt 2 )^5} = {3^5} + {5.3^4}.\sqrt 2 + {10.3^3}. {(\sqrt 2 )^2} + {10.3^2}. {(\sqrt 2 )^3} + 5.3. {(\sqrt 2 )^4} + {(\sqrt 2 )^5}{(3 – \sqrt 2 )^5}\\ = {3^5} – {5.3^4}.\sqrt 2 + {10.3^3}. {(\sqrt 2 )^2} – {10.3^2}. {(\sqrt 2 )^3} + 5.3. {(\sqrt 2 )^4} – {(\sqrt 2 )^5}\\ \Rightarrow {(3 + \sqrt 2 )^5} – {(3 – \sqrt 2 )^5} = 810\sqrt 2 + 360\sqrt 2 + 8\sqrt 2 = 1178\sqrt 2 \end{array}$$

### Solution 8.15 page 75 Math textbook 10 Connecting knowledge volume 2

a) Use the first two terms in the expansion of (1 + 0.02)5 to calculate the approximate value of 1.025

b) Use a calculator to calculate the value of 1.025and calculate the absolute error of the approximation obtained in question a.

Solution method

The formulas that expand $${\left( {a + b} \right)^n}$$ with $$n \in \left\{ {4;5} \right\}$$, are a tool. effective for calculating or approximating some quantity without the use of a calculator.

Detailed explanation

a) 1.025 =(1 +0.02)5 $$\approx$$ 15 + 5.14.0.02 = 1.1

b) We have: |1.025 – 1.1| < 0.0005

The absolute error is 0.0005.

### Solve problem 8.16 page 75 Math textbook 10 Connecting knowledge volume 2

The population of a province at the present time is about 800,000 people. Assume that the annual population growth rate of that province is r%.

a) Write the formula to calculate the population of that province after 1 year, after 2 years. From that, the formula to calculate the population of that province after 5 years is $$P=800\left ( 1+\frac{r}{100} \right )^{5}$$ (thousand people).

b) For r = 1.5%, use the first two terms in the expansion of (1 + 0.015)5 Estimate the population of that province after 5 years (in thousands of people).

Solution method

Calculate the population of that province after 1 year is: $$P_{1}=800 + \frac{r}{100}.800$$

Calculate the population of that province after 2 years is: $$P_{2}=P_{1}+ \frac{r}{100}.P_{1}$$

Derive the population of that province after 5 years

Detailed explanation

a) The population of that province after 1 year is: $$P_{1}=800 + \frac{r}{100}.800 = 800\left ( 1+\frac{r}{100} \right )\ ) (thousand people). The population of that province after 2 years is: \(P_{2}=P_{1}+ \frac{r}{100}.P_{1} = P_{1}.(1+ \frac{r}{ 100}) = 800\left ( 1+\frac{r}{100} \right ).(1+ \frac{r}{100}) = 800\left ( 1+\frac{r}{100} \ right )^{2}$$ (thousand people).

The population of that province after 5 years is: $$P_{5}= 800\left ( 1+\frac{r}{100} \right )^{5}$$ (thousand people).

b) (1 + 0.015)5 $$\approx 1^{5}+5.1^{4}.0.015=1,075$$

The population of that province after 5 years is approximately: 800,1,075 = 860 thousand people.