Answer :
Certainly! Let's go through the problem step-by-step.
### 7. Find the width of the park, given the area and the length:
#### Given:
- Area of the park: [tex]\(A = (x^2 - 19x + 90) \, \text{m}^2\)[/tex]
- Length of the park: [tex]\(L = (x - 9) \, \text{m}\)[/tex]
We need to find the width [tex]\(W\)[/tex] of the park.
#### Solution:
The area of a rectangular park is given by:
[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]
So, we have:
[tex]\[ x^2 - 19x + 90 = (x - 9) \times \text{Width} \][/tex]
To find the width, we divide the area by the length:
[tex]\[ \text{Width} = \frac{x^2 - 19x + 90}{x - 9} \][/tex]
Simplifying the division:
[tex]\[ \frac{x^2 - 19x + 90}{x - 9} = x - 10 \][/tex]
Thus, the width of the park is:
[tex]\[ W = x - 10 \, \text{m} \][/tex]
### 6. Find the quotient and the remainder:
#### (i) [tex]\(\frac{5x^3 + x - 3}{x - 1}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(5x^3\)[/tex] by the leading term of the denominator [tex]\(x\)[/tex]:
[tex]\[ \frac{5x^3}{x} = 5x^2 \][/tex]
2. Multiply [tex]\(5x^2 \cdot (x - 1) = 5x^3 - 5x^2\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (5x^3 + x - 3) - (5x^3 - 5x^2) = 5x^2 + x - 3 \][/tex]
4. Repeat the steps for the resulting polynomial [tex]\(5x^2 + x - 3\)[/tex].
Continuing the division, we get:
- Quotient: [tex]\(5x^2 + 5x + 6\)[/tex]
- Remainder: [tex]\(3\)[/tex]
So,
[tex]\[ \frac{5x^3 + x - 3}{x - 1} = 5x^2 + 5x + 6 + \frac{3}{x - 1} \][/tex]
#### (ii) [tex]\(\frac{x^3 + 8x^2 - 9}{x + 7}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(x^3\)[/tex] by the leading term of the denominator [tex]\(x\)[/tex]:
[tex]\[ \frac{x^3}{x} = x^2 \][/tex]
2. Multiply [tex]\(x^2 \cdot (x + 7) = x^3 + 7x^2\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (x^3 + 8x^2 - 9) - (x^3 + 7x^2) = x^2 - 9 \][/tex]
4. Repeat the steps for the resulting polynomial [tex]\(x^2 - 9\)[/tex].
Continuing the division, we get:
- Quotient: [tex]\(x^2 + x - 1\)[/tex]
- Remainder: [tex]\(-16\)[/tex]
So,
[tex]\[ \frac{x^3 + 8x^2 - 9}{x + 7} = x^2 + x - 1 - \frac{16}{x + 7} \][/tex]
#### (iii) [tex]\(\frac{14x^2 - 29xy + 18y^2}{2x - 5y}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(14x^2\)[/tex] by the leading term of the denominator [tex]\(2x\)[/tex]:
[tex]\[ \frac{14x^2}{2x} = 7x \][/tex]
2. Multiply [tex]\(7x \cdot (2x - 5y) = 14x^2 - 35xy\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (14x^2 - 29xy + 18y^2) - (14x^2 - 35xy) = 6xy + 18y^2 \][/tex]
4. Divide [tex]\(6xy\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[ \frac{6xy}{2x} = 3y \][/tex]
5. Multiply [tex]\(3y \cdot (2x - 5y) = 6xy - 15y^2\)[/tex]
6. Subtract the result from the previous polynomial:
[tex]\[ (6xy + 18y^2) - (6xy - 15y^2) = 33y^2 \][/tex]
Continuing the division, we get:
- Quotient: [tex]\(7x + 3y\)[/tex]
- Remainder: [tex]\(33y^2\)[/tex]
So,
[tex]\[ \frac{14x^2 - 29xy + 18y^2}{2x - 5y} = 7x + 3y + \frac{33y^2}{2x - 5y} \][/tex]
#### (iv) [tex]\(\frac{6x^2 + 15x + 9}{2x + 3}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(6x^2\)[/tex] by the leading term of the denominator [tex]\(2x\)[/tex]:
[tex]\[ \frac{6x^2}{2x} = 3x \][/tex]
2. Multiply [tex]\(3x \cdot (2x + 3) = 6x^2 + 9x\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (6x^2 + 15x + 9) - (6x^2 + 9x) = 6x + 9 \][/tex]
4. Divide [tex]\(6x\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[ \frac{6x}{2x} = 3 \][/tex]
5. Multiply [tex]\(3 \cdot (2x + 3) = 6x + 9\)[/tex]
6. Subtract the result from the previous polynomial:
[tex]\[ (6x + 9) - (6x + 9) = 0 \][/tex]
So,
- Quotient: [tex]\(3x + 3\)[/tex]
- Remainder: [tex]\(0\)[/tex]
Thus,
[tex]\[ \frac{6x^2 + 15x + 9}{2x + 3} = 3x + 3\][/tex]
### 7. Find the width of the park, given the area and the length:
#### Given:
- Area of the park: [tex]\(A = (x^2 - 19x + 90) \, \text{m}^2\)[/tex]
- Length of the park: [tex]\(L = (x - 9) \, \text{m}\)[/tex]
We need to find the width [tex]\(W\)[/tex] of the park.
#### Solution:
The area of a rectangular park is given by:
[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]
So, we have:
[tex]\[ x^2 - 19x + 90 = (x - 9) \times \text{Width} \][/tex]
To find the width, we divide the area by the length:
[tex]\[ \text{Width} = \frac{x^2 - 19x + 90}{x - 9} \][/tex]
Simplifying the division:
[tex]\[ \frac{x^2 - 19x + 90}{x - 9} = x - 10 \][/tex]
Thus, the width of the park is:
[tex]\[ W = x - 10 \, \text{m} \][/tex]
### 6. Find the quotient and the remainder:
#### (i) [tex]\(\frac{5x^3 + x - 3}{x - 1}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(5x^3\)[/tex] by the leading term of the denominator [tex]\(x\)[/tex]:
[tex]\[ \frac{5x^3}{x} = 5x^2 \][/tex]
2. Multiply [tex]\(5x^2 \cdot (x - 1) = 5x^3 - 5x^2\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (5x^3 + x - 3) - (5x^3 - 5x^2) = 5x^2 + x - 3 \][/tex]
4. Repeat the steps for the resulting polynomial [tex]\(5x^2 + x - 3\)[/tex].
Continuing the division, we get:
- Quotient: [tex]\(5x^2 + 5x + 6\)[/tex]
- Remainder: [tex]\(3\)[/tex]
So,
[tex]\[ \frac{5x^3 + x - 3}{x - 1} = 5x^2 + 5x + 6 + \frac{3}{x - 1} \][/tex]
#### (ii) [tex]\(\frac{x^3 + 8x^2 - 9}{x + 7}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(x^3\)[/tex] by the leading term of the denominator [tex]\(x\)[/tex]:
[tex]\[ \frac{x^3}{x} = x^2 \][/tex]
2. Multiply [tex]\(x^2 \cdot (x + 7) = x^3 + 7x^2\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (x^3 + 8x^2 - 9) - (x^3 + 7x^2) = x^2 - 9 \][/tex]
4. Repeat the steps for the resulting polynomial [tex]\(x^2 - 9\)[/tex].
Continuing the division, we get:
- Quotient: [tex]\(x^2 + x - 1\)[/tex]
- Remainder: [tex]\(-16\)[/tex]
So,
[tex]\[ \frac{x^3 + 8x^2 - 9}{x + 7} = x^2 + x - 1 - \frac{16}{x + 7} \][/tex]
#### (iii) [tex]\(\frac{14x^2 - 29xy + 18y^2}{2x - 5y}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(14x^2\)[/tex] by the leading term of the denominator [tex]\(2x\)[/tex]:
[tex]\[ \frac{14x^2}{2x} = 7x \][/tex]
2. Multiply [tex]\(7x \cdot (2x - 5y) = 14x^2 - 35xy\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (14x^2 - 29xy + 18y^2) - (14x^2 - 35xy) = 6xy + 18y^2 \][/tex]
4. Divide [tex]\(6xy\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[ \frac{6xy}{2x} = 3y \][/tex]
5. Multiply [tex]\(3y \cdot (2x - 5y) = 6xy - 15y^2\)[/tex]
6. Subtract the result from the previous polynomial:
[tex]\[ (6xy + 18y^2) - (6xy - 15y^2) = 33y^2 \][/tex]
Continuing the division, we get:
- Quotient: [tex]\(7x + 3y\)[/tex]
- Remainder: [tex]\(33y^2\)[/tex]
So,
[tex]\[ \frac{14x^2 - 29xy + 18y^2}{2x - 5y} = 7x + 3y + \frac{33y^2}{2x - 5y} \][/tex]
#### (iv) [tex]\(\frac{6x^2 + 15x + 9}{2x + 3}\)[/tex]:
Perform polynomial long division:
1. Divide the leading term of the numerator [tex]\(6x^2\)[/tex] by the leading term of the denominator [tex]\(2x\)[/tex]:
[tex]\[ \frac{6x^2}{2x} = 3x \][/tex]
2. Multiply [tex]\(3x \cdot (2x + 3) = 6x^2 + 9x\)[/tex]
3. Subtract the result from the original polynomial:
[tex]\[ (6x^2 + 15x + 9) - (6x^2 + 9x) = 6x + 9 \][/tex]
4. Divide [tex]\(6x\)[/tex] by [tex]\(2x\)[/tex]:
[tex]\[ \frac{6x}{2x} = 3 \][/tex]
5. Multiply [tex]\(3 \cdot (2x + 3) = 6x + 9\)[/tex]
6. Subtract the result from the previous polynomial:
[tex]\[ (6x + 9) - (6x + 9) = 0 \][/tex]
So,
- Quotient: [tex]\(3x + 3\)[/tex]
- Remainder: [tex]\(0\)[/tex]
Thus,
[tex]\[ \frac{6x^2 + 15x + 9}{2x + 3} = 3x + 3\][/tex]
