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Solving Exercises Lesson 4: Multiplication and division of one-variable polynomials (C7 Math 7 Horizon)
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Solution 1 page 40 Math textbook 7 Creative horizon volume 2
Perform multiplication.
a) \((4x – 3)(x + 2)\)
b) \((5x + 2)( – {x^2} + 3x + 1)\)
c) \((2{x^2} – 7x + 4)( – 3{x^2} + 6x + 5)\)
Solution method
Apply the rules for multiplying polynomials (distributions).
Detailed explanation
a) \(\begin{array}{l}(4x – 3)(x + 2) = 4x(x + 2) – 3(x + 2)\\ = 4{x^2} + 8x – 3x – 6\end{array}\)
\( = 4{x^2} + 5x – 6\)
b) \((5x + 2)( – {x^2} + 3x + 1)\)
\( = 5x( – {x^2} + 3x + 1) + 2( – {x^2} + 3x + 1)\)
\( = – 5{x^3} + 15{x^2} + 5x – 2{x^2} + 6x + 2\)
\( = – 5{x^3} + 13{x^2} + 11x + 2\)
c) \((2{x^2} – 7x + 4)( – 3{x^2} + 6x + 5)\)
\( = 2{x^2}( – 3{x^2} + 6x + 5) – 7x( – 3{x^2} + 6x + 5) + 4( – 3{x^2} + 6x + 5)\)
\( = 2{x^2}( – 3{x^2}) + 2{x^2}.6x + 2{x^2}.5 + 7x.3{x^2} – 7x.6x – 7x.5 + 4( – 3{x^2}) + 4.6x + 4.5\)
\(= – 6{x^4} + 33{x^3} – 44{x^2} – 11x + 20\)
Solution 2 page 40 Math textbook 7 Creative horizon volume 2
Given two rectangles as shown in Figure 4. Find the polynomial in terms of the variable x representing the area of the blue shaded part
Solution method
We calculate the area of the large rectangle
We calculate the area of the small rectangle
Subtract the large area from the small area to find the area to find
Detailed explanation
The area of the large rectangle is: \((2x + 4)(3x + 2) = 2x + 4(3x + 2) = 6{x^2} + 4x + 12x + 8 = 6{x^2 } + 16x + 8\)
The area of the small rectangle is : \(x(x + 1) = {x^2} + x\)
The area to find is: \(6{x^2} + 16x + 8 – {x^2} – x\)\( = 5{x^2} + 15x + 8\)
Solve problems 3 pages 40 Math textbook 7 Creative horizon volume 2
Perform division.
a) \((8{x^6} – 4{x^5} + 12{x^4} – 20{x^3}):4{x^3}\)
b) \((2{x^2} – 5x + 3):(2x – 3)\)
Solution method
Perform multiplication and division of polynomials by setting the calculator.
Note: Sort polynomials in descending order of powers
Detailed explanation
a) \((8{x^6} – 4{x^5} + 12{x^4} – 20{x^3}):4{x^3}\)
\( = (8{x^6}:4{x^3}) – (4{x^5}:4{x^3}) + (12{x^4}:4{x^3}) – (20{x^3}:4{x^3})\)
\( = 2{x^2} – {x^2} + 3x – 5\)
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b)
So \((2{x^2} – 5x + 3):(2x – 3)= x – 1\)
Solve problems 4 pages 40 Math textbook 7 Creative horizon volume 2
Perform division.
a) \((4{x^2} – 5):(x – 2)\)
b) \((3{x^3} – 7x + 2):(2{x^2} – 3)\)
Solution method
Use learned rules to divide polynomials
It is recommended to arrange the polynomial in descending power to make it easy to calculate
Detailed explanation
a) \((4{x^2} – 5):(x – 2) = \dfrac{{4{x^2} – 5}}{{x – 2}} = 4x + 8 + \dfrac{ {11}}{{x – 2}}\)
So \( (4{x^2} – 5):(x – 2)= 4x + 8 + \dfrac{{11}}{{x – 2}}\)
b) \((3{x^3} – 7x + 2):(2{x^2} – 3) = \dfrac{{3{x^3} – 7x + 2}}{{2{x^ 2} – 3}}\)
So \( (3{x^3} – 7x + 2):(2{x^2} – 3)= \dfrac{3}{2}x – \dfrac{{\dfrac{5}{2}x + 2}}{{2{x^2} – 3}}\)
Solve problems 5 pages 40 Math textbook 7 Creative horizon volume 2
Calculate the length of a rectangle whose area is equal to \((4{y^2} + 4y – 3)\)\(c{m^2}\) and width is equal to (2y – 1) cm.
Solution method
Use the formula to calculate the area of a rectangle
Apply the polynomial division rule and arrange the polynomials in descending order of powers
Detailed explanation
Area of rectangle = length. width
\( \Rightarrow (4{y^2} + 4y – 3):(2y – 1) = \dfrac{{4{y^2} + 4y – 3}}{{2y – 1}}\)
So the length of the rectangle is: 2y + 3 cm
Solving problems 6 pages 40 Math textbook 7 Creative horizon volume 2
Given a rectangular box with volume equal to (\(3{x^3} + 8{x^2} – 45x – 50\)) \(c{m^3}\), length equal to (x + 5 ) cm and the height is equal to (x + 1) cm. Calculate the width of that rectangular box.
Solution method
Use the formula to calculate the volume of a rectangular box V=Sh
Apply the rule of division by two polynomials
Detailed explanation
The volume of a rectangular box is equal to the product of the height and the area of the base
So width of rectangle = volume : ( height . length )
The area of the base is \((x + 5)(x + 1) = {x^2} + 6x + 5\)
Substituting the numbers in the problem for the above formula we get:
\( = \dfrac{{3{x^3} + 8{x^2} – 45x – 50}}{{(x + 5)(x + 1) = {x^2} + 6x + 5}} = \dfrac{{3{x^3} + 8{x^2} – 45x – 50}}{{{x^2} + 6x + 5}}\)
So the length of the rectangular box is 3x – 10 cm
