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SBT Solution Lesson 2: One-variable polynomial (C7 SBT Math 7 Horizon)

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### Solve problem 1 page 27 SBT Math 7 Creative horizon episode 2 – CTST

Which of the following expressions is a one-variable polynomial?

\(A = – 4\); \(B = 2t + 9\); \(C = \frac{{3x – 4}}{{2x + 1}}\); \(N = \frac{{1 – 2y}}{3}\); \(M = 4 + 7y – 2{y^3}\)

**Detailed instructions for solving Lesson 1**

**Solution method**

Understand the concept of one-variable monomial, one-variable polynomial to determine.

A monomial of a variable is an algebraic expression consisting of only a number, or a variable, or a product between numbers and the variable.

A one-variable polynomial is the sum of all mononomials of the same variable.

**Detailed explanation**

We have \(N = \frac{{1 – 2y}}{3} = \frac{1}{3} – \frac{2}{3}y\).

Hence the one-variable polynomials are:

\(A = – 4\); \(B = 2t + 9\); \(N = \frac{{1 – 2y}}{3}\); \(M = 4 + 7y – 2{y^3}\)

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### Solve problem 2 page 27 SBT Math 7 Creative horizon episode 2 – CTST

Let the polynomial \(P\left( x \right) = 3{x^2} + 8{x^3} – 2x + 4{x^3} – 2{x^2} + 9\). Arrange the monomials according to the decreasing power of the variable.

**Detailed instructions for solving Lesson 2**

**Solution method**

Step 1: Add and subtract mononomials of the same variable to reduce the given polynomial.

Step 2: Sort the monomials by the descending power of the variable.

**Detailed explanation**

\(P\left( x \right) = 3{x^2} + 8{x^3} – 2x + 4{x^3} – 2{x^2} + 9 = 12{x^3} + {x^2} – 2x + 9\)

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### Solve problem 3 page 27 SBT Math 7 Creative horizon episode 2 – CTST

Let the polynomial \(P\left( x \right) = 4{x^2} + 2{x^3} – 15x + 7{x^3} – 9{x^2} + 6 + 5x\). State the degree, the highest coefficient and the coefficient of freedom of the polynomial \(P\left( x \right)\).

**Detailed instructions for solving Lesson 3**

**Solution method**

Step 1: Simplify the polynomial.

Step 2: Based on the concepts of order, highest coefficient, and coefficient of freedom of the polynomial to answer.

The degree of a one-variable polynomial (non-zero polynomial) is the largest exponent of the variable in that polynomial when in reduced form.

The highest coefficient is the coefficient of the monomial with the highest degree in the polynomial.

The coefficient of freedom is the coefficient that does not contain the variable x.

**Detailed explanation**

We have \(P\left( x \right) = 4{x^2} + 2{x^3} – 15x + 7{x^3} – 9{x^2} + 6 + 5x = 9{x ^3} – 5{x^2} – 10x + 6\)

The degree of the polynomial \(P\left( x \right)\) is 3.

The highest coefficient is 9.

The coefficient of freedom is 6.

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### Solve problem 4 page 27 SBT Math 7 Creative horizon episode 2 – CTST

Let’s calculate the value of the polynomials:

a) \(P\left( x \right) = – 3{x^3} + 8{x^2} – 2x + 1\) when \(x = – 3\).

b) \(Q\left( y \right) = 7{y^3} – 6{y^4} + 3{y^2} – 2y\) when \(y = 2\).

**Detailed instructions for solving Lesson 4**

**Solution method**

Step 1: Reduce the polynomial (if possible).

Step 2: Replace the variable’s values with numbers to calculate.

**Detailed explanation**

a) Replace \(x = – 3\) into \(P\left( x \right) = – 3{x^3} + 8{x^2} – 2x + 1\) we get \(P\left ( { – 3} \right) = – 3. {\left( { – 3} \right)^3} + 8. {\left( { – 3} \right)^2} – 2.\left( { – 3} \right) + 1 = 160\)

b) Replace \(y = 2\) into \(Q\left( y \right) = 7{y^3} – 6{y^4} + 3{y^2} – 2y\) we have \( Q\left( 2 \right) = {7.2^3} – {6.2^4} + {3.2^2} – 2.2 = – 32\).

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### Solve problem 5 page 27 SBT Math 7 Creative horizon episode 2 – CTST

Is \(x = – \frac{4}{5}\) a solution of \(P\left( x \right) = 5x + 4\)?

**Detailed instructions for solving Lesson 5**

**Solution method**

Replace \(x = {x_0}\) into \(P\left( x \right)\) if \(P\left( {{x_0}} \right) = 0\) then \(x = {x_0} \) is the solution of \(P\left( x \right)\).

**Detailed explanation**

Replace \(x = – \frac{4}{5}\) into \(P\left( x \right) = 5x + 4\), we get \(P\left( { – \frac{4}{) 5}} \right) = 5.\left( {\frac{{ – 4}}{5}} \right) + 4 = – 4 + 4 = 0\)

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So \(x = – \frac{4}{5}\) is a solution of \(P\left( x \right) = 5x + 4\).

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### Solve problem 6 page 27 SBT Math 7 Creative horizon episode 2 – CTST

Let the polynomial \(Q\left( t \right) = 3{t^2} + 15t + 12\). Indicate which numbers in the set \(\left\{ {1; – 4; – 1} \right\}\) are solutions of \(Q\left( t \right)\).

**Detailed instructions for solving Lesson 6**

**Solution method**

Replace \(t = {t_0}\) into \(Q\left( t \right)\) if \(Q\left( {{t_0}} \right) = 0\) then \(t = {t_0} \) is the solution of \(Q\left( t \right)\).

**Detailed explanation**

+ Replace \(t = 1\) into \(Q\left( t \right) = 3{t^2} + 15t + 12\), we have \(Q\left( 1 \right) = {3.1^ 2} + 15.1 + 12 = 30 \ne 0\)

So \(t = 1\) is not a solution of \(Q\left( t \right) = 3{t^2} + 15t + 12\).

+ Replace \(t = – 4\) into \(Q\left( t \right) = 3{t^2} + 15t + 12\), we have \(Q\left( { – 4} \right) = 3. {\left( { – 4} \right)^2} + 15.\left( { – 4} \right) + 12 = 0\)

So \(t = – 4\) is the solution of \(Q\left( t \right) = 3{t^2} + 15t + 12\).

+ Replace \(t = – 1\) into \(Q\left( t \right) = 3{t^2} + 15t + 12\), we have \(Q\left( { – 1} \right) = 3. {\left( { – 1} \right)^2} + 15.\left( { – 1} \right) + 12 = 0\)

So \(t = – 1\) is the solution of \(Q\left( t \right) = 3{t^2} + 15t + 12\).

So \(\left\{ { – 4; – 1} \right\}\) is the solution of \(Q\left( t \right) = 3{t^2} + 15t + 12\).

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### Solution 7 page 28 SBT Math 7 Creative horizon episode 2 – CTST

Does the polynomial \(M\left( t \right) = – 8 – 3{t^2}\) have a solution? Why?

**Detailed instructions for solving Lesson 7**

**Solution method**

Use the knowledge you have learned about powers to compare the given polynomial with zero.

**Detailed explanation**

We have \({t^2} \ge 0;\,\,\forall t\)

Derive \( – 3{t^2} \le 0 \Rightarrow – 8 – 3{t^2} \le – 8\) or \(M\left( t \right) \le – 8\)

The maximum value of \(M\left( t \right) = – 8\) so the polynomial \(M\left( t \right) = – 8 – 3{t^2}\) has no solution.

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### Solve problem 8 page 28 SBT Math 7 Creative horizon episode 2 – CTST

In volleyball, a tee shot can be described by the expression \(h = – 4.9{t^2} + 3.8t + 1.6\), where h is the height of the ball. sso is measured in meters and t is the time since tee-off in seconds. Calculate the height h when \(t = 0.4\)seconds.

**Detailed instructions for solving Lesson 8**

**Solution method**

Substitute the value of the variable into the given expression to calculate.

**Detailed explanation**

Replace \(t = 0.4\) into \(h = – 4.9{t^2} + 3.8t + 1.6\); we have \(h = – 4,9.0,{4^2} + 3.8.0.4 + 1.6 = 2,336\) meters.

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### Solve problem 9 page 28 SBT Math 7 Creative horizon episode 2 – CTST

Consider a rectangular garden whose perimeter is \(80\) meters and length is x meters. Write an expression that represents the area of the garden. Calculate the area of the garden when \(x = 25\)m.

**Detailed instructions for solving Lesson 9**

**Solution method**

Step 1: Express other quantities in terms of the given unknowns.

Step 2: Represent the relationship between the quantities in the implicit by mathematical operations.

**Detailed explanation**

Half the perimeter of the garden is \(80:2 = 40\) m.

The width of the garden is \(80 – x\) m.

The area of the garden is \(x.\left( {80 – x} \right)\) m^{2}

When \(x = 25\) we have \(25.\left( {80 – 25} \right) = 1375\) m^{2}.

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### Solution 10 page 28 SBT Math 7 Creative horizon episode 2 – CTST

The height of a firework above the ground is described by the expression \(h = – 4.8{t^2} + 21.6t + 156\), where h is in meters and t is the time since the shot is in seconds (only consider \(0 < t < 2.2\)). Calculate the height h when \(t = 2\) seconds.

**Detailed instructions for solving Lesson 10**

**Solution method**

Substitute the value of the variable into the given expression to calculate.

**Detailed explanation**

Substituting \(t = 2\)seconds into \(h = – 4.8{t^2} + 21.6t + 156\) we have

\(h = – 4,{8.2^2} + 21,6.2 + 156 = 180\) meters.

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