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Solving Exercises Lesson 25 One-variable Polynomials (Chapter 7 Math 7 Connection)
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Solve problem 7.5 page 30 Math textbook 7 Connecting knowledge volume 2 – KNTT
a) Calculate \(\left( {\dfrac{1}{2}{x^3}} \right).\left( {4{x^2}} \right)\). Find the coefficient and degree of the obtained monomial.
b) Calculate \(\dfrac{1}{2}{x^3} – \dfrac{5}{2}{x^3}\). Find the coefficient and degree of the obtained monomial.
Detailed instructions for solving Problem 7.5
Solution method
Step 1: Collapse
a) To multiply two mononomials, we multiply the two coefficients together and multiply the two powers of the variable together
b) To subtract two mononomials of the same degree, we subtract the coefficients together, keeping the power of the variable.
Step 2:
The monomial takes the form of the product of a real number to a power of the variable.
The real number is called the coefficient
The exponent of the power of the variable is called the degree of the monomial
Detailed explanation
a) \(\left( {\dfrac{1}{2}{x^3}} \right).\left( {4{x^2}} \right) = \left( {\dfrac{1} {2}.4} \right).\left( {{x^3}. {x^2}} \right) = 2. {x^5}\).
Coefficient: 2
Grade: 5
b) \(\dfrac{1}{2}{x^3} – \dfrac{5}{2}{x^3} = \left( {\dfrac{1}{2} – \dfrac{5} {2}} \right){x^3} = \dfrac{{ – 4}}{2}. {x^3} = – 2{x^3}\)
Coefficient: -2
Grade: 3
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Solve lesson 7.6, page 30, Math 7 textbook Connecting knowledge volume 2 – KNTT
Given two polynomials:
\(\begin{array}{l}A = {x^3} + \dfrac{3}{2}x – 7{x^4} + \dfrac{1}{2}x – 4{x^2 } + 9\\B = {x^5} – 3{x^2} + 8{x^4} – 5{x^2} – {x^5} + x – 7\end{array}\)
a) Collapse and sort the above two polynomials according to the decreasing power of the variable.
b) Find the degree, highest coefficient and coefficient of freedom for each given polynomial.
Detailed instructions for solving problems 7.6
Solution method
a) Step 1: Add and subtract mononomials of the same degree to get a reduced polynomial that contains no two mononomials of the same degree
Step 2: Sort the above polynomial according to the decreasing power of the variable.
b) + Degree of the polynomial is the degree of the term with the highest degree
+ The highest coefficient is the coefficient of the term with the highest degree
+ The coefficient of freedom is the coefficient of the term 0.
Detailed explanation
a)
\(\begin{array}{l}A(x) = {x^3} + \dfrac{3}{2}x – 7{x^4} + \dfrac{1}{2}x – 4{ x^2} + 9\\ = – 7{x^4} + {x^3} – 4{x^2} + \left( {\dfrac{3}{2}x + \dfrac{1}{ 2}x} \right) + 9\\ = – 7{x^4} + {x^3} – 4{x^2} + 2x + 9\\B(x) = {x^5} – 3 {x^2} + 8{x^4} – 5{x^2} – {x^5} + x – 7\\ = \left( {{x^5} – {x^5}} \right ) + 8{x^4} + \left( { – 3{x^2} – 5{x^2}} \right) + x – 7\\ = 0 + 8{x^4} + ( – 8 {x^2}) + x – 7\\ = 8{x^4} – 8{x^2} + x – 7\end{array}\)
b) * Polynomial A(x):
+ Degree of the polynomial is: 4
+ The highest coefficient is: -7
+ The coefficient of freedom is: 9
* Polynomial B(x):
+ Degree of the polynomial is: 4
+ The highest coefficient is: 8
+ The coefficient of freedom is: -7
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Solving problem 7.7, page 30, Math 7 Textbook Connecting knowledge volume 2 – KNTT
Given two polynomials:
\(\begin{array}{l}P(x) = 5{x^3} + 2{x^4} – {x^2} + 3{x^2} – {x^3} – 2{ x^4} – 4{x^3}\\Q(x) = 3x – 4{x^3} + 8{x^2} – 5x + 4{x^3} + 5\end{array}\ )
a) Collapse and sort the above two polynomials according to the decreasing power of the variable.
b) Find the degree, highest coefficient and coefficient of freedom for each given polynomial.
Detailed instructions for solving Problem 7.7
Solution method
a) Step 1: Add and subtract mononomials of the same degree to get a reduced polynomial that contains no two mononomials of the same degree
Step 2: Sort the above polynomial according to the decreasing power of the variable.
b) Replace each x value into P(x), Q(x) has been reduced and calculated.
Detailed explanation
a)
\(\begin{array}{l}P(x) = 5{x^3} + 2{x^4} – {x^2} + 3{x^2} – {x^3} – 2{ x^4} – 4{x^3}\\ = \left( {2{x^4} – 2{x^4}} \right) + \left( {5{x^3} – {x^ 3} – 4{x^3}} \right) + \left( { – {x^2} + 3{x^2}} \right)\\ = 0 + 0 + 2{x^2}\\ = 2{x^2}\\Q(x) = 3x – 4{x^3} + 8{x^2} – 5x + 4{x^3} + 5\\ = \left( { – 4{ x^3} + 4{x^3}} \right) + 8{x^2} + \left( {3x – 5x} \right) + 5\\ = 0 + 8{x^2} + ( – 2x) + 5\\ = 8{x^2} – 2x + 5\end{array}\)
b) P(1) = 2.12 = 2
P(0) = 2.02 = 0
Q(-1) = 8.(-1)2 – 2.(-1) +5 = 8 +2 +5 =15
Q(0) = 8.02 – 2.0 + 5 = 5
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Solve problem 7.8, page 30 Math 7 Textbook Connecting knowledge volume 2 – KNTT
Two pumps are used to pump water into a water tank. The first machine pumps 22 m . per hour3 water. The second pump pumps 16 m . per hour3 water. After both machines run for x hours, the first machine is turned off and the second machine runs for another 0.5 hours, the water tank is full.
Write the polynomial (x variable) representing the tank capacity (m .)3). Know that before pumping, in the tank there are 1.5 m3 water. Find the highest coefficient and the coefficient of freedom of that polynomial.
Detailed instructions for solving Problem 7.8
Solution method
Step 1: Write a polynomial expressing tank capacity = Amount of water 2 pumps in x hours + amount of tap water 2 pumps in 0.5 hours + Amount of water available in the tank
Step 2: Collapse the polynomial
+ The highest coefficient is the coefficient of the term with the highest degree
+ The coefficient of freedom is the coefficient of the term 0.
Detailed explanation
Polynomial V(x) = 22.x + 16.x + 0.5.16 + 1.5 = (22+16).x + 8 + 1.5 = 38.x + 9.5
Highest coefficient: 38
Factor of freedom: 9.5
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Solve problem 7.9 page 30 Math textbook 7 Connecting knowledge volume 2 – KNTT
Write a polynomial F(x) that satisfies the following conditions at the same time:
The degree of F(x) is 3
The coefficient of x2 is equal to the factor of x and is equal to 2
The highest coefficient of F(x) is -6 and the free factor is 3.
Detailed instructions for solving problem 7.9
Solution method
Write a polynomial that satisfies the following requirements:
The degree of a polynomial is the degree of the term with the highest degree
+ The highest coefficient is the coefficient of the term with the highest degree
+ The coefficient of freedom is the coefficient of the term 0.
Detailed explanation
F(x) = -6x3 + 2x2 + 2x + 3
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Solution 7.10, page 30 Math 7 Textbook Connecting knowledge volume 2 – KNTT
Give it a check:
a) \(x = – \dfrac{1}{8}\) is the solution of the polynomial P(x) = 4x + \(\dfrac{1}{2}\)?
b) Of the three numbers 1; -1 and 2, which number is the solution of the polynomial Q(x) = x2 + x – 2 ?
Detailed instructions for solving problems 7.10
Solution method
a) Replace the value \(x = – \dfrac{1}{8}\) into the polynomial P(x) = 4x + \(\dfrac{1}{2}\) to calculate the value P(\( – \dfrac{1}{8}\)). If P(\( – \dfrac{1}{8}\)) = 0, then \(x = – \dfrac{1}{8}\) is a solution of P(x)
b) Find Q(1); Q(-1); Q(2). At what value of x does Q(x) = 0 then that number is the solution of Q(x)
Detailed explanation
a) We have: P(\( – \dfrac{1}{8}\)) = 4.(\( – \dfrac{1}{8}\))+ \(\dfrac{1}{2} \)= (-\(\dfrac{1}{2}\)) + \(\dfrac{1}{2}\) = 0
So \(x = – \dfrac{1}{8}\) is the solution of the polynomial P(x) = 4x + \(\dfrac{1}{2}\)
b) Q(1) = 12 +1 – 2 = 0
Q(-1) = (-1)2 + (-1) – 2 = -2
Q(2) = 22 + 2 – 2 = 4
Since Q(1) = 0, x = 1 is the solution of Q(x)
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Solve problem 7.11 page 30 Math textbook 7 Connecting knowledge volume 2 – KNTT
Mother gave Quynh 100 thousand dong. Quynh bought a set of learning tools for 37,000 VND and a Math reference book for x (thousand VND).
a) Find the polynomial (variable x) representing the amount of Quynh remaining (unit: thousand VND). Find the degree of that polynomial.
b) After buying the book, Quynh has just spent all the money her mother gave her. What is the price of the book?
Detailed instructions for solving problem 7.11
Solution method
Write a polynomial that represents the remaining amount = the amount given by the mother – the amount purchased
Degree of a polynomial is the degree of the term with the highest degree
When all the money is spent, the remaining amount is equal to 0
Detailed explanation
a) Polynomial C(x) = 100 – 37 – x = – x + 63
Degree of polynomial is 1
b) After buying the book, we have 0 money left or – x + 63 = 0
\( \Rightarrow 63 = x\) or x = 63
So the price of the book is 63,000 dong
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