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**SBT Prize End of Chapter 7 – Math 7 SBT Horizon**

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### Solution 1 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Let \(B = x{y^3} + 4xy – 2{x^2} + 3\). Calculate the value of the expression \(B\) when \(x = – 1\), \(y = 2\).

**Detailed instructions for solving Lesson 1**

**Solution method**

Substitute the values of the variable and calculate.

**Detailed explanation**

Substituting \(x = – 1\), \(y = 2\) into \(B = x{y^3} + 4xy – 2{x^2} + 3\) we have

\(B = \left( { – 1} \right){.2^3} + 4\left( { – 1} \right).2 – 2. {\left( { – 1} \right)^2 } + 3 = – 15\).

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### Solution 2 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Which of the following expressions is a monomial of one variable?

a) \(2y\);

b) \(3x + 5\);

c) \(12\);

d) \(\frac{1}{3}{t^2}\).

**Detailed instructions for solving Lesson 2**

**Solution method**

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.

**Detailed explanation**

The one-variable monomial is a; c; d.

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### Solution 3 page 33 SBT Math 7 Creative horizon episode 2 – CTST

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

\(5 – 2x\); \(6{x^2} + 8{x^3} + 3x – 2\);

\(\frac{2}{{x – 1}}\); \(\frac{1}{4}t – 5\).

**Detailed instructions for solving Lesson 3**

**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**

The one-variable polynomials are “ \(5 – 2x;\,\,6{x^2} + 8{x^3} + 3x – 2\); \(\frac{1}{4}t – 5\).

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### Solution 4 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Write a quaternary one-variable polynomial with 5 terms.

**Detailed instructions for solving Lesson 4**

**Solution method**

A quaternary one-variable polynomial means that the degree of the largest variable is 4.

**Detailed explanation**

A quaternary one-variable polynomial with 5 terms is:

A(x) = x^{4} – 2x^{3} + 3x^{2} – 4x + 5.

* Comment: *This post has many answers.

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### Solve problem 5 page 33 SBT Math 7 Creative horizon episode 2 – CTST

State the degrees of the following polynomials:

\(A = 5{x^2} – 2{x^4} + 7\);

\(B = 17\);

\(C = 3x – 4{x^3} + 2{x^2} + 1\).

**Detailed instructions for solving Lesson 5**

**Solution method**

The degree of the polynomial is the maximum exponent of the variable after reducing the polynomial.

**Detailed explanation**

The degree of \(A = 5{x^2} – 2{x^4} + 7\) is 2.

The degree of \(B = 17\) is 0.

The degree of \(C = 3x – 4{x^3} + 2{x^2} + 1\) is 3.

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### Solution 6 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Let the polynomial \(P\left( x \right) = {x^3} + 64\). Find the root of the polynomial in the set \(\left\{ {0;4; – 4} \right\}\)

**Detailed instructions for solving Lesson 6**

**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**

With \(x = 0\) we have \(P\left( 0 \right) = {0^3} + 64 = 64 \ne 0\), so \(0\) is not a solution of \(P \left( x \right)\).

With \(x = 4\) we have \(P\left( 4 \right) = {4^3} + 64 = 128 \ne 0\), so \(4\) is not a solution of \(P \left( x \right)\).

With \(x = – 4\) we have \(P\left( { – 4} \right) = {\left( { – 4} \right)^3} + 64 = 0\), deduce \( – 4\) is the solution of \(P\left( x \right)\).

So \( – 4\) is the solution of \(P\left( x \right)\).

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### Solution 7 page 33 SBT Math 7 Creative horizon episode 2 – CTST

The length of the triangle is \(3y + 2\); \(6y – 4\) and the perimeter is \(23y – 5\). Find the unknown side in that triangle.

**Detailed instructions for solving Lesson 7**

**Solution method**

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The perimeter of a triangle is equal to the sum of its three sides. So, if we want to find the unknown edge, we subtract the known edges from the perimeter.

**Detailed explanation**

The unknown side in that triangle is \(23y – 5 – \left( {3y + 2} \right) – \left( {6y – 4} \right) = 14y – 3\).

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### Solve problem 8 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Let the polynomial \(M\left( x \right) = 3{x^5} – 4{x^3} + 9x + 2\). Find the polynomials \(N\left( x \right)\), \(Q\left( x \right)\) such that \(N\left( x \right) – M\left( x \right) = – 5{x^4} – 4{x^3} + 2{x^2} + 8x\) and \(Q\left( x \right) + M\left( x \right) = 3{x ^4} – 2{x^3} + 9{x^2} – 7\)

**Detailed instructions for solving Lesson 8**

**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.

Step 3: Perform the calculation horizontally or vertically.

**Detailed explanation**

\(N\left( x \right) – M\left( x \right) = – 5{x^4} – 4{x^3} + 2{x^2} + 8x \Rightarrow N\left( x \right) = – 5{x^4} – 4{x^3} + 2{x^2} + 8x + M\left( x \right)\)

\( – 5{x^4} – 4{x^3} + 2{x^2} + 8x + 3{x^5} – 4{x^3} + 9x + 2 = 3{x^5} – 5{x^4} – 8{x^3} + 2{x^2} + 17x + 2\)

So \(N\left( x \right) = 3{x^5} – 5{x^4} – 8{x^3} + 2{x^2} + 17x + 2\).

\(Q\left( x \right) + M\left( x \right) = 3{x^4} – 2{x^3} + 9{x^2} – 7 \Rightarrow Q\left( x \ right) = 3{x^4} – 2{x^3} + 9{x^2} – 7 – M\left( x \right)\)

\(3{x^4} – 2{x^3} + 9{x^2} – 7 – \left( {3{x^5} – 4{x^3} + 9x + 2} \right) = 3{x^4} – 2{x^3} + 9{x^2} – 7 – 3{x^5} + 4{x^3} – 9x – 2 = – 3{x^5} + 3{x^4} + 2{x^3} + 9{x^2} – 9x – 9\)

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### Solve lesson 9 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform multiplication.

a) \(\left( {4x – 5} \right)\left( {3x + 4} \right)\);

b) \(\left( {2{x^2} – 3x + 5} \right)\left( {4x + 3} \right)\)

**Detailed instructions for solving Lesson 9**

**Solution method**

Understand the rule of one-variable polynomials: To multiply a polynomial by a polynomial, we multiply each term of one polynomial by each term of the other polynomial and then add the products together.

**Detailed explanation**

a) \(\left( {4x – 5} \right)\left( {3x + 4} \right) = 4x.3x + 4x.4 – 5.3x – 5.4 = 12{x^2} + x – 20 \);

b) \(\left( {2{x^2} – 3x + 5} \right)\left( {4x + 3} \right) = 2{x^2}.4x + 2{x^2}. 3 – 3x.4x – 3x.3 + 5.4x + 5.3 = 8{x^3} – 6{x^2} + 11x + 15\)

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### Solution 10 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform division.

a) \(\left( {64{y^2} – 16{y^4} + 8{y^5}} \right):4y\)

b) \(\left( {5{t^2} – 8t + 3} \right):\left( {t – 1} \right)\)

**Detailed instructions for solving Lesson 10**

**Solution method**

**Detailed explanation**

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### Solve lesson 11 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform division.

a) \(\left( {{x^4} + 6{x^2} + 8} \right):\left( {{x^2} + 2} \right)\)

b) \(\left( {3{x^3} – 2{x^2} + 3x – 2} \right):\left( {{x^2} + 1} \right)\)

**Detailed instructions for solving Lesson 11**

**Solution method**

Divide polynomials divide polynomials by setting the division calculation.

**Detailed explanation**

a)

So \(\left( {{x^4} + 6{x^2} + 8} \right):\left( {{x^2} + 2} \right) = {x^2} + 4\ )

b)

So \(\left( {3{x^3} – 2{x^2} + 3x – 2} \right):\left( {{x^2} + 1} \right) = 3x – 2\) .

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### Solution 12 page 34 SBT Math 7 Creative horizon episode 2 – CTST

Perform division.

a) \(\left( {2{x^2} – 7x + 4} \right):\left( {x – 2} \right)\)

b) \(\left( {2{x^3} + 3{x^2} + 3x + 4} \right):\left( {{x^2} + 2} \right)\)

**Detailed instructions for solving Lesson 12**

**Solution method**

Divide polynomials divide polynomials by setting the division calculation.

**Detailed explanation**

a)

So \(\frac{{2{x^2} – 7x + 4}}{{x – 2}} = 2x – 3 – \frac{2}{{x – 2}}\).

b)

So \(\frac{{2{x^3} + 3{x^2} + 3x + 4}}{{{x^2} + 2}} = 2x + 3 – \frac{{x + 2} }{{{x^2} + 2}}\)

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