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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.
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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.
