Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To solve the given expression, let's break it down step-by-step.
### Step 1: Factorize Each Polynomial
1. [tex]\( \frac{3x^2 + 2x - 21}{-2x^2 - 2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49} \)[/tex]
#### Numerator and Denominator of the First Fraction
Numerator: [tex]\(3x^2 + 2x - 21\)[/tex]
- We need to factorize [tex]\(3x^2 + 2x - 21\)[/tex].
Denominator: [tex]\(-2x^2 - 2x + 12\)[/tex]
- We need to factorize [tex]\(-2x^2 - 2x + 12\)[/tex]. We can start by factoring out [tex]\(-2\)[/tex].
#### Numerator and Denominator of the Second Fraction
Numerator: [tex]\(2x^2 + 25x + 63\)[/tex]
- We need to factorize [tex]\(2x^2 + 25x + 63\)[/tex].
Denominator: [tex]\(6x^2 + 7x - 49\)[/tex]
- We need to factorize [tex]\(6x^2 + 7x - 49\)[/tex].
### Step 2: Rewrite in Factored Form
After factorizing:
1. [tex]\(3x^2 + 2x - 21 = (3x - 7)(x + 3)\)[/tex]
2. [tex]\(-2x^2 - 2x + 12 = -2(x^2 + x - 6) = -2(x - 2)(x + 3)\)[/tex]
3. [tex]\(2x^2 + 25x + 63 = (2x + 9)(x + 7)\)[/tex]
4. [tex]\(6x^2 + 7x - 49 = (3x + 7)(2x - 7)\)[/tex]
Combining all these factorizations, we get:
[tex]\[ \frac{(3x - 7)(x + 3)}{-2(x - 2)(x + 3)} \cdot \frac{(2x + 9)(x + 7)}{(3x + 7)(2x - 7)} \][/tex]
### Step 3: Cancel Common Factors
- Cancel out the common factors in the numerator and denominator.
[tex]\[ \frac{(\cancel{3x - 7})(\cancel{x + 3})}{-2(\cancel{x - 2})(\cancel{x + 3})} \cdot \frac{(2x + 9)(x + 7)}{(\cancel{3x + 7})(2x - 7)} \][/tex]
- Simplify the remaining terms:
[tex]\[ \frac{(2x + 9)(x + 7)}{-2(2x - 7)} \][/tex]
### Step 4: Combine the Fractions
- Multiply the numerators and denominators together:
[tex]\[ \frac{(2x + 9)(x + 7)}{-2(2x - 7)} \][/tex]
### Given [tex]\(a = 1\)[/tex]:
- Suppose our simplified fraction needs to match the form [tex]\( \frac{ax + b}{cx + d} \)[/tex].
If [tex]\(a = 1\)[/tex], then:
[tex]\[ \frac{1x + b}{cx + d} = \frac{(2x + 9)(x + 7)}{-2(2x - 7)} \][/tex]
By observing the forms, we compare the coefficients:
If [tex]\(a = 1\)[/tex]:
- The equivalent expression is found by matching terms:
[tex]\[ b = 16 \][/tex]
[tex]\[ c = -2 \][/tex]
[tex]\[ d = -14 \][/tex]
### Answer
[tex]\[ b = 16 \][/tex]
[tex]\[ c = -2 \][/tex]
[tex]\[ d = -14 \][/tex]
### Step 1: Factorize Each Polynomial
1. [tex]\( \frac{3x^2 + 2x - 21}{-2x^2 - 2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49} \)[/tex]
#### Numerator and Denominator of the First Fraction
Numerator: [tex]\(3x^2 + 2x - 21\)[/tex]
- We need to factorize [tex]\(3x^2 + 2x - 21\)[/tex].
Denominator: [tex]\(-2x^2 - 2x + 12\)[/tex]
- We need to factorize [tex]\(-2x^2 - 2x + 12\)[/tex]. We can start by factoring out [tex]\(-2\)[/tex].
#### Numerator and Denominator of the Second Fraction
Numerator: [tex]\(2x^2 + 25x + 63\)[/tex]
- We need to factorize [tex]\(2x^2 + 25x + 63\)[/tex].
Denominator: [tex]\(6x^2 + 7x - 49\)[/tex]
- We need to factorize [tex]\(6x^2 + 7x - 49\)[/tex].
### Step 2: Rewrite in Factored Form
After factorizing:
1. [tex]\(3x^2 + 2x - 21 = (3x - 7)(x + 3)\)[/tex]
2. [tex]\(-2x^2 - 2x + 12 = -2(x^2 + x - 6) = -2(x - 2)(x + 3)\)[/tex]
3. [tex]\(2x^2 + 25x + 63 = (2x + 9)(x + 7)\)[/tex]
4. [tex]\(6x^2 + 7x - 49 = (3x + 7)(2x - 7)\)[/tex]
Combining all these factorizations, we get:
[tex]\[ \frac{(3x - 7)(x + 3)}{-2(x - 2)(x + 3)} \cdot \frac{(2x + 9)(x + 7)}{(3x + 7)(2x - 7)} \][/tex]
### Step 3: Cancel Common Factors
- Cancel out the common factors in the numerator and denominator.
[tex]\[ \frac{(\cancel{3x - 7})(\cancel{x + 3})}{-2(\cancel{x - 2})(\cancel{x + 3})} \cdot \frac{(2x + 9)(x + 7)}{(\cancel{3x + 7})(2x - 7)} \][/tex]
- Simplify the remaining terms:
[tex]\[ \frac{(2x + 9)(x + 7)}{-2(2x - 7)} \][/tex]
### Step 4: Combine the Fractions
- Multiply the numerators and denominators together:
[tex]\[ \frac{(2x + 9)(x + 7)}{-2(2x - 7)} \][/tex]
### Given [tex]\(a = 1\)[/tex]:
- Suppose our simplified fraction needs to match the form [tex]\( \frac{ax + b}{cx + d} \)[/tex].
If [tex]\(a = 1\)[/tex], then:
[tex]\[ \frac{1x + b}{cx + d} = \frac{(2x + 9)(x + 7)}{-2(2x - 7)} \][/tex]
By observing the forms, we compare the coefficients:
If [tex]\(a = 1\)[/tex]:
- The equivalent expression is found by matching terms:
[tex]\[ b = 16 \][/tex]
[tex]\[ c = -2 \][/tex]
[tex]\[ d = -14 \][/tex]
### Answer
[tex]\[ b = 16 \][/tex]
[tex]\[ c = -2 \][/tex]
[tex]\[ d = -14 \][/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.