Answer :
Sure, let's break down and simplify each of the given expressions step by step.
### Expression 1: [tex]\(\frac{x^2-6x-7}{x^2-10x+21}\)[/tex]
1. Factor both the numerator and the denominator:
- [tex]\(x^2 - 6x - 7\)[/tex] can be factored as [tex]\((x - 7)(x + 1)\)[/tex]
- [tex]\(x^2 - 10x + 21\)[/tex] can be factored as [tex]\((x - 7)(x - 3)\)[/tex]
2. Write the factored form:
[tex]\[ \frac{(x - 7)(x + 1)}{(x - 7)(x - 3)} \][/tex]
3. Cancel the common factor [tex]\((x - 7)\)[/tex] from the numerator and denominator:
[tex]\[ \frac{x + 1}{x - 3} \][/tex]
So, the simplified form of the first expression is [tex]\(\frac{x + 1}{x - 3}\)[/tex].
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### Expression 2: [tex]\(2 \cdot \frac{x^2+6x+9}{x^2-9}\)[/tex]
1. Factor both the numerator and the denominator:
- [tex]\(x^2 + 6x + 9\)[/tex] can be factored as [tex]\((x + 3)^2\)[/tex]
- [tex]\(x^2 - 9\)[/tex] can be factored as [tex]\((x + 3)(x - 3)\)[/tex] (difference of squares)
2. Write the factored form:
[tex]\[ 2 \cdot \frac{(x + 3)^2}{(x + 3)(x - 3)} \][/tex]
3. Cancel the common factor [tex]\((x + 3)\)[/tex] from the numerator and denominator:
[tex]\[ 2 \cdot \frac{x + 3}{x - 3} \][/tex]
So, the simplified form of the second expression is [tex]\(2 \cdot \frac{x + 3}{x - 3}\)[/tex].
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### Expression 3: [tex]\(\frac{2x^2 - x - 28}{20 - x - x^2}\)[/tex]
1. Factor both the numerator and the denominator:
- [tex]\(2x^2 - x - 28\)[/tex] can be factored as [tex]\((2x + 7)(x - 4)\)[/tex]
- [tex]\(20 - x - x^2\)[/tex] can be rearranged as [tex]\( - (x^2 + x - 20) \)[/tex]. Factoring [tex]\(x^2 + x - 20\)[/tex] results in [tex]\((x + 5)(x - 4)\)[/tex].
2. Rewriting the denominator with a negative sign:
[tex]\[ 20 - x - x^2 = -(x^2 + x - 20) = -(x + 5)(x - 4) \][/tex]
3. Write the factored form:
[tex]\[ \frac{(2x + 7)(x - 4)}{-(x + 5)(x - 4)} \][/tex]
4. Cancel the common factor [tex]\((x - 4)\)[/tex] from the numerator and denominator:
[tex]\[ \frac{(2x + 7)}{-(x + 5)} = \frac{-(2x + 7)}{(x + 5)} = \frac{-2x - 7}{x + 5} \][/tex]
So, the simplified form of the third expression is [tex]\(\frac{-2x - 7}{x + 5}\)[/tex].
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In conclusion, the simplified forms of the expressions are:
1. [tex]\(\frac{x + 1}{x - 3}\)[/tex]
2. [tex]\(2 \cdot \frac{x + 3}{x - 3}\)[/tex]
3. [tex]\(\frac{-2x - 7}{x + 5}\)[/tex]
### Expression 1: [tex]\(\frac{x^2-6x-7}{x^2-10x+21}\)[/tex]
1. Factor both the numerator and the denominator:
- [tex]\(x^2 - 6x - 7\)[/tex] can be factored as [tex]\((x - 7)(x + 1)\)[/tex]
- [tex]\(x^2 - 10x + 21\)[/tex] can be factored as [tex]\((x - 7)(x - 3)\)[/tex]
2. Write the factored form:
[tex]\[ \frac{(x - 7)(x + 1)}{(x - 7)(x - 3)} \][/tex]
3. Cancel the common factor [tex]\((x - 7)\)[/tex] from the numerator and denominator:
[tex]\[ \frac{x + 1}{x - 3} \][/tex]
So, the simplified form of the first expression is [tex]\(\frac{x + 1}{x - 3}\)[/tex].
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### Expression 2: [tex]\(2 \cdot \frac{x^2+6x+9}{x^2-9}\)[/tex]
1. Factor both the numerator and the denominator:
- [tex]\(x^2 + 6x + 9\)[/tex] can be factored as [tex]\((x + 3)^2\)[/tex]
- [tex]\(x^2 - 9\)[/tex] can be factored as [tex]\((x + 3)(x - 3)\)[/tex] (difference of squares)
2. Write the factored form:
[tex]\[ 2 \cdot \frac{(x + 3)^2}{(x + 3)(x - 3)} \][/tex]
3. Cancel the common factor [tex]\((x + 3)\)[/tex] from the numerator and denominator:
[tex]\[ 2 \cdot \frac{x + 3}{x - 3} \][/tex]
So, the simplified form of the second expression is [tex]\(2 \cdot \frac{x + 3}{x - 3}\)[/tex].
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### Expression 3: [tex]\(\frac{2x^2 - x - 28}{20 - x - x^2}\)[/tex]
1. Factor both the numerator and the denominator:
- [tex]\(2x^2 - x - 28\)[/tex] can be factored as [tex]\((2x + 7)(x - 4)\)[/tex]
- [tex]\(20 - x - x^2\)[/tex] can be rearranged as [tex]\( - (x^2 + x - 20) \)[/tex]. Factoring [tex]\(x^2 + x - 20\)[/tex] results in [tex]\((x + 5)(x - 4)\)[/tex].
2. Rewriting the denominator with a negative sign:
[tex]\[ 20 - x - x^2 = -(x^2 + x - 20) = -(x + 5)(x - 4) \][/tex]
3. Write the factored form:
[tex]\[ \frac{(2x + 7)(x - 4)}{-(x + 5)(x - 4)} \][/tex]
4. Cancel the common factor [tex]\((x - 4)\)[/tex] from the numerator and denominator:
[tex]\[ \frac{(2x + 7)}{-(x + 5)} = \frac{-(2x + 7)}{(x + 5)} = \frac{-2x - 7}{x + 5} \][/tex]
So, the simplified form of the third expression is [tex]\(\frac{-2x - 7}{x + 5}\)[/tex].
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In conclusion, the simplified forms of the expressions are:
1. [tex]\(\frac{x + 1}{x - 3}\)[/tex]
2. [tex]\(2 \cdot \frac{x + 3}{x - 3}\)[/tex]
3. [tex]\(\frac{-2x - 7}{x + 5}\)[/tex]