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Solving exercises at the end of CHAPTER 7: Algebraic Expressions – Math 7 Horizons
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Solve problem 1 page 42 Math textbook 7 Creative horizon volume 2
Let \(A = {x^2}y + 2xy – 3{y^2} + 4\). Calculate the value of expression A when x = -2, y = 3.
Solution method
We replace the given x, y problems into the expression and calculate
Detailed explanation
\(A = {x^2}y + 2xy – 3{y^2} + 4\)
Substituting x = -2 and y = 3 into the formula we have:
\(\begin{array}{l}A = {( – 2)^2}.3 + 2( – 2).3 – {3.3^2} + 4\\ = 4.3 – 12 – 27 + 4\\ = – 23\end{array}\)
Solution 2 page 42 Math textbook 7 Creative horizon volume 2
Which of the following expressions is a monomial of one variable?
a) 2y b) 3x + 5
c) 8 d)\(21{t^{12}}\)
Solution method
One-variable polynomial definition.
Detailed explanation
The one-variable polynomials are a, c, d
Solution 3 page 42 Math textbook 7 Creative horizon volume 2
Which of the following expressions is a one-variable polynomial?
\(3 + 6y\);
\(7{x^2} + 2x – 4{x^4} + 1\);
\(\dfrac{2}{{x + 1}}\);
\(\dfrac{1}{3}x – 5\).
Solution method
Based on the definition of a one-variable polynomial
Detailed explanation
The one-variable polynomials are:
\(3 + 6y;7{x^2} + 2x – 4{x^4} + 1;\dfrac{1}{3}x – 5\)
Solution 4 page 42 Math textbook 7 Creative horizon volume 2
Write a cubic one-variable polynomial with 3 terms.
Solution method
Using degree definitions in 1-variable polynomials
Detailed explanation
\({x^3} + 2x – 1\)
Attention : There are different ways to write polynomials but in this lesson the terms in the polynomial are always 3
Solve problems 5 pages 42 Math textbook 7 Creative horizon volume 2
Give the degrees of the following polynomials:
\(A = 3x – 4{x^2} + 1\)
\(B = 7\)
\(M = x – 7{x^3} + 10{x^4} + 2\)
Solution method
Based on the definitions of degrees in polynomials
Detailed explanation
A has degree 2
B has degree 0
M has degree 4
Solve lesson 6 page 42 Math textbook 7 Creative horizon volume 2
Let the polynomial P(x) = \({x^3} + 27\). Find the solution of P(x) in the set \(\left\{ {0;3; – 3} \right\}\)
Solution method
We consider P(x) = 0 and then find x. The value x found is the solution of the polynomial
Detailed explanation
Consider P(x) = \({x^3} + 27 = 0\)
\(\begin{array}{l} \Leftrightarrow {x^3} = – 27\\ \Leftrightarrow {x^3} = – 27 = {( – 3)^3}\\ \Rightarrow x = – 3\ end{array}\)
Since \( – 3 \in \left\{ {0;3; – 3} \right\}\) -3 is a solution
Solution 7 page 42 Math textbook 7 Creative horizon volume 2
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The triangle in Figure 1 has a perimeter of (25y – 8) cm. Find the unknown side in that triangle.
Solution method
We calculate the remaining side by subtracting the other 2 known sides from the perimeter
Use the rule of addition and subtraction of polynomials
Detailed explanation
According to the problem, we have the perimeter of the triangle = 25y – 8 cm
We have 2 sides of a known triangle according to the problem
\( \Rightarrow \) The remaining side to be found of the triangle is: 25y – 8 – 5y + 3 – 7y + 4 = 13y – 7 cm
Solve problem 8 page 42 Math textbook 7 Creative horizon volume 2
Let the polynomial \(M(x) = 2{x^4} – 5{x^3} + 7{x^2} + 3x\).
Find the polynomials N(x), Q(x) such that:
\(N(x) – M(x) = – 4{x^4} – 2{x^3} + 6{x^2} + 7\)
and \(M(x) + Q(x) = 6{x^5} – {x^4} + 3{x^2} – 2\)
Solution method
Apply the rule of addition and subtraction of 1-variable polynomials
Detailed explanation
According to the problem we have \(M(x) = 2{x^4} – 5{x^3} + 7{x^2} + 3x\)
\(\begin{array}{l}M(x) + Q(x) = 6{x^5} – {x^4} + 3{x^2} – 2\\ \Rightarrow Q(x) = (6{x^5} – {x^4} + 3{x^2} – 2) – (2{x^4} – 5{x^3} + 7{x^2} + 3x)\\ \Rightarrow Q(x) = 6{x^5} – {x^4} + 3{x^2} – 2 – 2{x^4} + 5{x^3} – 7{x^2} – 3x\\Q(x) = 6{x^5} – 3{x^4} + 5{x^3} – 4{x^2} – 3x – 2\end{array}\)
According to the title we have:
\(\begin{array}{l}N(x) – M(x) = – 4{x^4} – 2{x^3} + 6{x^2} + 7\\ \Rightarrow N(x ) = – 4{x^4} – 2{x^3} + 6{x^2} + 7 + 2{x^4} – 5{x^3} + 7{x^2} + 3x\\ \Rightarrow N(x) = – 2{x^4} – 7{x^3} + 13{x^2} + 3x + 7\end{array}\)
Solve lesson 9 page 42 Math textbook 7 Creative horizon volume 2
Perform multiplication.
a) \((3x – 2)(4x + 5)\)
b) \(({x^2} – 5x + 4)(6x + 1)\)
Solution method
Apply the rule for multiplying 2 polynomials with 1 variable
Detailed explanation
a) \((3x – 2)(4x + 5)\)
\(\begin{array}{l} = 3x(4x + 5) – 2(4x + 5)\\ = 3x.4x + 5.3x – 2.4x – 2.5\\ = 12{x^2} + 7x – 10\end{array}\)
b) \(({x^2} – 5x + 4)(6x + 1)\)
\(\begin{array}{l} = {x^2}(6x + 1) – 5x(6x + 1) + 4(6x + 1)\\ = {x^2}.6x + 1. {x ^2} – 5x.6x – 5x.1 + 4.6x + 4.1\end{array}\)
\( = 6{x^3} – 29{x^2} + 19x + 4\)
Solve problems 10 pages 42 Math textbook 7 Creative horizon volume 2
Perform division.
a) \((45{x^5} – 5{x^4} + 10{x^2}):5{x^2}\)
b) \((9{t^2} – 3{t^4} + 27{t^5}):3t\)
Solution method
Perform the calculation using the polynomial division rule
We can divide by columns, but we need to sort the polynomials in descending order of powers
Detailed explanation
a) \((45{x^5} – 5{x^4} + 10{x^2}):5{x^2}\)\( = 9{x^3} – {x^2} + 2\)
b) \((9{t^2} – 3{t^4} + 27{t^5}):3t = (27{t^5} – 3{t^4} + 9{t^2} ):3t\\=(27t^5):(3t) – (3t^4):(3t)+(9t^2):(3t) = 9{t^4} – 3{t^3}+ 3 T\)
Solve problem 11 page 42 Math textbook 7 Creative horizon volume 2
Perform division.
a) \((2{y^4} – 13{y^3} + 15{y^2} + 11y – 3):({y^2} – 4y – 3)\)
b) \((5{x^3} – 3{x^2} + 10):({x^2} + 1)\)
Solution method
Calculating and dividing 2 polynomials
We arrange the polynomials in descending order of powers to make the calculation easier
Detailed explanation
\(a)(2{y^4} – 13{y^3} + 15{y^2} + 11y – 3):({y^2} – 4y – 3)=2y^2-5y+1 \)
b) \((5{x^3} – 3{x^2} + 10):({x^2} + 1)=5x-3+\dfrac{-5x+13}{x^2+1 }\)
