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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]
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