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**Solving SBT Lesson 4: Multiplication and division of one-variable polynomials (C7 SBT Math 7 Horizons)**

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### Solve problem 1 page 32 SBT Math 7 Creative horizon episode 2 – CTST

Multiply \(\left( {7x – 2} \right)\left( { – 2x + 5} \right)\).

**Detailed instructions for solving Lesson 1**

**Solution method**

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

**Detailed explanation**

\(\left( {7x – 2} \right)\left( { – 2x + 5} \right) = 7x\left( { – 2x} \right) + 7x.5 + \left( { – 2} \ right)\left( { – 2x} \right) + \left( { – 2} \right).5 = – 14{x^2} + 39x – 10\)

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### Solution 2 page 32 SBT Math 7 Creative horizon episode 2 – CTST

Multiply \(\left( {3x – 4} \right)\left( { – 2{x^2} + 7x + 4} \right)\).

**Detailed instructions for solving Lesson 2**

**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**

\(\begin{array}{l}\left( {3x – 4} \right)\left( { – 2{x^2} + 7x + 4} \right)\\ = 3x\left( { – 2 {x^2}} \right) + 3x.7x + 3x.4 + \left( { – 4} \right).\left( { – 2{x^2}} \right) + \left( { – 4} \right).7x + \left( { – 4} \right).4\\ = – 6{x^3} + 21{x^2} + 12x + 8{x^2} – 28x – 16 \\ = – 6{x^3} + 29{x^2} – 16x – 16\end{array}\)

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### Solution 3 page 32 SBT Math 7 Creative horizon episode 2 – CTST

Multiply \(\left( {4{x^2} – 2x + 1} \right)\left( { – 2{x^2} + 5x + 3} \right)\).

**Detailed instructions for solving Lesson 3**

**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**

\(\begin{array}{l}\left( {4{x^2} – 2x + 1} \right)\left( { – 2{x^2} + 5x + 3} \right)\\ = 4{x^2}\left( { – 2{x^2}} \right) + 4{x^2}.5x + 4{x^2}.3 – 2x.\left( { – 2{x ^2}} \right) – 2x.5x – 2x.3 + \left( { – 2{x^2}} \right) + 5x + 3\\ = – 8{x^4} + 20{x^ 3} + 12{x^2} + 4{x^3} – 10{x^2} – 6x – 2{x^2} + 5x + 3\\ = – 8{x^4} + 24{x ^3} – x + 3\end{array}\)

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

Make a polynomial expression in the variable \(x\) representing the area of the shaded part in Figure 1.

**Detailed instructions for solving Lesson 4**

**Solution method**

Use the learned area formulas to write the expression.

Area of a rectangle is the product of length and width

**Detailed explanation**

The area of the large rectangle is \(\left( {3x + 2} \right)\left( {2x + 4} \right) = 2x.3x + 2x.2 + 4.3x + 4.2 = 6{x^2 } + 16x + 8\).

The area of the small rectangle is \(x\left( {x + 1} \right) = {x^2} + x\).

The required area is \(6{x^2} + 16x + 8 – \left( {{x^2} + x} \right) = 5{x^2} + 15x + 8\).

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

Do the division \(\left( {9{x^5} – 15{x^4} + 27{x^3} – 12{x^2}} \right):3{x^2}\) .

**Detailed instructions for solving Lesson 5**

**Solution method**

Polynomial division by monomial: Divide each term of the polynomial by the monomial and add the quotients together.

**Detailed explanation**

\(\begin{array}{l}\left( {9{x^5} – 15{x^4} + 27{x^3} – 12{x^2}} \right):3{x^ 2}\\ = 9{x^5}:3{x^2} + \left( { – 15{x^4}} \right):3{x^2} + 27{x^3}:3 {x^2} + \left( { – 12{x^2}} \right):3{x^2}\\ = 3{x^3} – 5{x^2} + 9x – 4\end {array}\)

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

Do the division \(\left( {2{x^2} – 5x + 3} \right):\left( {2x – 3} \right)\).

**Detailed instructions for solving Lesson 6**

**Solution method**

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Divide polynomials divide polynomials by setting the division calculation.

**Detailed explanation**

So \(\left( {2{x^2} – 5x + 3} \right):\left( {2x – 3} \right) = x – 1\)

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

Do the division \(\left( {4{x^2} – 5} \right):\left( {x – 2} \right)\).

**Detailed instructions for solving Lesson 7**

**Solution method**

Divide polynomials divide polynomials by setting the division calculation.

**Detailed explanation**

We perform polynomial division:

So \(\frac{{4{x^2} – 5}}{{x – 2}} = 4x + 8 + \frac{{11}}{{x – 2}}\).

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### Solve problem 8 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Do the division \(\left( {4{x^3} – 7x + 2} \right):\left( {2{x^2} – 3} \right)\).

**Detailed instructions for solving lesson 8**

**Solution method**

Divide polynomials divide polynomials by setting the division calculation.

**Detailed explanation**

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

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### Solve lesson 9 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Calculate the length of a rectangle whose area is \(4{y^2} + 4y – 3\) (cm^{2}) and the width is equal to \(\left( {2y – 1} \right)\)(cm).

**Detailed instructions for solving Lesson 9**

**Solution method**

The length of the rectangle is the quotient of the area divided by the width.

**Detailed explanation**

We have

So the length of the given rectangle is \(2y + 3\) cm.

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### Solution 10 page 33 SBT Math 7 Creative horizon episode 2 – CTST

Given a rectangular box with volume \(V = 3{x^3} + 8{x^2} – 45x – 50\) (cm^{3}), length is equal to \(\left( {x + 5} \right)\) cm and height \(\left( {x + 1} \right)\) cm. Calculate the width of the rectangular box.

**Detailed instructions for solving Lesson 10**

**Solution method**

The volume of a rectangular box is the product of the three dimensions.

Width equals volume divided by product of length and height

**Detailed explanation**

We have the width calculated by the calculation:

\(\left( {3{x^3} + 8{x^2} – 45x – 50} \right):\left[ {\left( {x + 5} \right)\left( {x + 1} \right)} \right] = \left( {3{x^3} + 8{x^2} – 45x – 50} \right):\left( {{x^2} + 6x + 5} \right)\)

We have

So the width of the rectangle is \(3x – 10\) cm.

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