Solved on Mar 13, 2024

Simplify and express the given rational expression in the form $\frac{f(x)}{g(x)}$ .
Problem: Simplify $\frac{1}{x^{2}+3 x+2}-\frac{1}{x^{2}-3 x-10}$ and give your answers for $f(x)$ and $g(x)$ .

STEP 1

Assumptions
1. The expression to simplify is $\frac{1}{x^{2}+3x+2}-\frac{1}{x^{2}-3x-10}$ .
2. We aim to express the simplified form as $\frac{f(x)}{g(x)}$ .

STEP 2

Factor the denominators of both fractions to find common factors if any.
The first denominator is $x^{2}+3x+2$ and the second is $x^{2}-3x-10$ .

STEP 3

Factor the first denominator:
$x^{2}+3x+2 = (x+a)(x+b)$
where $a$ and $b$ are numbers that add up to 3 and multiply to 2.

STEP 4

Find the numbers $a$ and $b$ for the first denominator:
$a = 1, \quad b = 2$
because $1 + 2 = 3$ and $1 \times 2 = 2$ .

STEP 5

Write the factored form of the first denominator:
$x^{2}+3x+2 = (x+1)(x+2)$

STEP 6

Factor the second denominator:
$x^{2}-3x-10 = (x+c)(x+d)$
where $c$ and $d$ are numbers that add up to $-3$ and multiply to $-10$ .

STEP 7

Find the numbers $c$ and $d$ for the second denominator:
$c = -5, \quad d = 2$
because $-5 + 2 = -3$ and $-5 \times 2 = -10$ .

STEP 8

Write the factored form of the second denominator:
$x^{2}-3x-10 = (x-5)(x+2)$

STEP 9

Rewrite the original expression with the factored denominators:
$\frac{1}{(x+1)(x+2)} - \frac{1}{(x-5)(x+2)}$

STEP 10

Notice that the denominators have a common factor of $(x+2)$ . To combine the fractions, we need a common denominator, which will be the product of the distinct factors from both denominators.

STEP 11

Determine the common denominator:
$Common\, denominator = (x+1)(x+2)(x-5)$

STEP 12

Rewrite both fractions with the common denominator:
$\frac{1 \cdot (x-5)}{(x+1)(x+2)(x-5)} - \frac{1 \cdot (x+1)}{(x+1)(x+2)(x-5)}$

STEP 13

Combine the numerators over the common denominator:
$\frac{(x-5) - (x+1)}{(x+1)(x+2)(x-5)}$

STEP 14

Expand the numerators and combine like terms:
$\frac{x - 5 - x - 1}{(x+1)(x+2)(x-5)}$

STEP 15

Simplify the numerator:
$\frac{-6}{(x+1)(x+2)(x-5)}$

STEP 16

The simplified form of the expression is now:
$\frac{-6}{(x+1)(x+2)(x-5)}$

STEP 17

Write the final answer in the form $\frac{f(x)}{g(x)}$ :
$f(x) = -6$ $g(x) = (x+1)(x+2)(x-5)$
Your answer for the function $f(x)$ is $-6$ . Your answer for the function $g(x)$ is $(x+1)(x+2)(x-5)$ .

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Simplify and express the given rational expression in the form $\frac{f(x)}{g(x)}$ .
Problem: Simplify $\frac{1}{x^{2}+3 x+2}-\frac{1}{x^{2}-3 x-10}$ and give your answers for $f(x)$ and $g(x)$ .

STEP 1

Assumptions
1. The expression to simplify is $\frac{1}{x^{2}+3x+2}-\frac{1}{x^{2}-3x-10}$ .
2. We aim to express the simplified form as $\frac{f(x)}{g(x)}$ .

STEP 2

Factor the denominators of both fractions to find common factors if any.
The first denominator is $x^{2}+3x+2$ and the second is $x^{2}-3x-10$ .

STEP 3

Factor the first denominator:
$x^{2}+3x+2 = (x+a)(x+b)$
where $a$ and $b$ are numbers that add up to 3 and multiply to 2.

STEP 4

Find the numbers $a$ and $b$ for the first denominator:
$a = 1, \quad b = 2$
because $1 + 2 = 3$ and $1 \times 2 = 2$ .

STEP 5

Write the factored form of the first denominator:
$x^{2}+3x+2 = (x+1)(x+2)$

STEP 6

Factor the second denominator:
$x^{2}-3x-10 = (x+c)(x+d)$
where $c$ and $d$ are numbers that add up to $-3$ and multiply to $-10$ .

STEP 7

Find the numbers $c$ and $d$ for the second denominator:
$c = -5, \quad d = 2$
because $-5 + 2 = -3$ and $-5 \times 2 = -10$ .

STEP 8

Write the factored form of the second denominator:
$x^{2}-3x-10 = (x-5)(x+2)$

STEP 9

Rewrite the original expression with the factored denominators:
$\frac{1}{(x+1)(x+2)} - \frac{1}{(x-5)(x+2)}$

STEP 10

Notice that the denominators have a common factor of $(x+2)$ . To combine the fractions, we need a common denominator, which will be the product of the distinct factors from both denominators.

STEP 11

Determine the common denominator:
$Common\, denominator = (x+1)(x+2)(x-5)$

STEP 12

Rewrite both fractions with the common denominator:
$\frac{1 \cdot (x-5)}{(x+1)(x+2)(x-5)} - \frac{1 \cdot (x+1)}{(x+1)(x+2)(x-5)}$

STEP 13

Combine the numerators over the common denominator:
$\frac{(x-5) - (x+1)}{(x+1)(x+2)(x-5)}$

STEP 14

Expand the numerators and combine like terms:
$\frac{x - 5 - x - 1}{(x+1)(x+2)(x-5)}$

STEP 15

Simplify the numerator:
$\frac{-6}{(x+1)(x+2)(x-5)}$

STEP 16

The simplified form of the expression is now:
$\frac{-6}{(x+1)(x+2)(x-5)}$

STEP 17

Write the final answer in the form $\frac{f(x)}{g(x)}$ :
$f(x) = -6$ $g(x) = (x+1)(x+2)(x-5)$
Your answer for the function $f(x)$ is $-6$ . Your answer for the function $g(x)$ is $(x+1)(x+2)(x-5)$ .

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Simplify and express the given rational expression in the form f(x)g(x)\frac{f(x)}{g(x)}g(x)f(x)​.Problem: Simplify 1x2+3x+2−1x2−3x−10\frac{1}{x^{2}+3 x+2}-\frac{1}{x^{2}-3 x-10}x2+3x+21​−x2−3x−101​ and give your answers for f(x)f(x)f(x) and g(x)g(x)g(x).

STEP 1

Assumptions1. The expression to simplify is 1x2+3x+2−1x2−3x−10\frac{1}{x^{2}+3x+2}-\frac{1}{x^{2}-3x-10}x2+3x+21​−x2−3x−101​.2. We aim to express the simplified form as f(x)g(x)\frac{f(x)}{g(x)}g(x)f(x)​.

STEP 2

Factor the denominators of both fractions to find common factors if any.The first denominator is x2+3x+2x^{2}+3x+2x2+3x+2 and the second is x2−3x−10x^{2}-3x-10x2−3x−10.

STEP 3

Factor the first denominator:x2+3x+2=(x+a)(x+b)x^{2}+3x+2 = (x+a)(x+b)x2+3x+2=(x+a)(x+b)where aaa and bbb are numbers that add up to 3 and multiply to 2.

STEP 4

Find the numbers aaa and bbb for the first denominator:a=1,b=2a = 1, \quad b = 2a=1,b=2because 1+2=31 + 2 = 31+2=3 and 1×2=21 \times 2 = 21×2=2.

STEP 5

Write the factored form of the first denominator:x2+3x+2=(x+1)(x+2)x^{2}+3x+2 = (x+1)(x+2)x2+3x+2=(x+1)(x+2)

STEP 6

Factor the second denominator:x2−3x−10=(x+c)(x+d)x^{2}-3x-10 = (x+c)(x+d)x2−3x−10=(x+c)(x+d)where ccc and ddd are numbers that add up to −3-3−3 and multiply to −10-10−10.

STEP 7

Find the numbers ccc and ddd for the second denominator:c=−5,d=2c = -5, \quad d = 2c=−5,d=2because −5+2=−3-5 + 2 = -3−5+2=−3 and −5×2=−10-5 \times 2 = -10−5×2=−10.

STEP 8

Write the factored form of the second denominator:x2−3x−10=(x−5)(x+2)x^{2}-3x-10 = (x-5)(x+2)x2−3x−10=(x−5)(x+2)

STEP 9

Rewrite the original expression with the factored denominators:1(x+1)(x+2)−1(x−5)(x+2)\frac{1}{(x+1)(x+2)} - \frac{1}{(x-5)(x+2)}(x+1)(x+2)1​−(x−5)(x+2)1​

STEP 10

Notice that the denominators have a common factor of (x+2)(x+2)(x+2). To combine the fractions, we need a common denominator, which will be the product of the distinct factors from both denominators.

STEP 11

Determine the common denominator:Common denominator=(x+1)(x+2)(x−5)Common\, denominator = (x+1)(x+2)(x-5)Commondenominator=(x+1)(x+2)(x−5)

STEP 12

Rewrite both fractions with the common denominator:1⋅(x−5)(x+1)(x+2)(x−5)−1⋅(x+1)(x+1)(x+2)(x−5)\frac{1 \cdot (x-5)}{(x+1)(x+2)(x-5)} - \frac{1 \cdot (x+1)}{(x+1)(x+2)(x-5)}(x+1)(x+2)(x−5)1⋅(x−5)​−(x+1)(x+2)(x−5)1⋅(x+1)​

STEP 13

Combine the numerators over the common denominator:(x−5)−(x+1)(x+1)(x+2)(x−5)\frac{(x-5) - (x+1)}{(x+1)(x+2)(x-5)}(x+1)(x+2)(x−5)(x−5)−(x+1)​

STEP 14

Expand the numerators and combine like terms:x−5−x−1(x+1)(x+2)(x−5)\frac{x - 5 - x - 1}{(x+1)(x+2)(x-5)}(x+1)(x+2)(x−5)x−5−x−1​

STEP 15

Simplify the numerator:−6(x+1)(x+2)(x−5)\frac{-6}{(x+1)(x+2)(x-5)}(x+1)(x+2)(x−5)−6​

STEP 16

The simplified form of the expression is now:−6(x+1)(x+2)(x−5)\frac{-6}{(x+1)(x+2)(x-5)}(x+1)(x+2)(x−5)−6​

STEP 17

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Simplify and express the given rational expression in the form $\frac{f(x)}{g(x)}$ .
Problem: Simplify $\frac{1}{x^{2}+3 x+2}-\frac{1}{x^{2}-3 x-10}$ and give your answers for $f(x)$ and $g(x)$ .

Assumptions
1. The expression to simplify is $\frac{1}{x^{2}+3x+2}-\frac{1}{x^{2}-3x-10}$ .
2. We aim to express the simplified form as $\frac{f(x)}{g(x)}$ .

Factor the denominators of both fractions to find common factors if any.
The first denominator is $x^{2}+3x+2$ and the second is $x^{2}-3x-10$ .

Factor the first denominator:
$x^{2}+3x+2 = (x+a)(x+b)$
where $a$ and $b$ are numbers that add up to 3 and multiply to 2.

Find the numbers $a$ and $b$ for the first denominator:
$a = 1, \quad b = 2$
because $1 + 2 = 3$ and $1 \times 2 = 2$ .

Write the factored form of the first denominator:
$x^{2}+3x+2 = (x+1)(x+2)$

Factor the second denominator:
$x^{2}-3x-10 = (x+c)(x+d)$
where $c$ and $d$ are numbers that add up to $-3$ and multiply to $-10$ .

Find the numbers $c$ and $d$ for the second denominator:
$c = -5, \quad d = 2$
because $-5 + 2 = -3$ and $-5 \times 2 = -10$ .

Write the factored form of the second denominator:
$x^{2}-3x-10 = (x-5)(x+2)$

Rewrite the original expression with the factored denominators:
$\frac{1}{(x+1)(x+2)} - \frac{1}{(x-5)(x+2)}$

Notice that the denominators have a common factor of $(x+2)$ . To combine the fractions, we need a common denominator, which will be the product of the distinct factors from both denominators.

Determine the common denominator:
$Common\, denominator = (x+1)(x+2)(x-5)$

Rewrite both fractions with the common denominator:
$\frac{1 \cdot (x-5)}{(x+1)(x+2)(x-5)} - \frac{1 \cdot (x+1)}{(x+1)(x+2)(x-5)}$

Combine the numerators over the common denominator:
$\frac{(x-5) - (x+1)}{(x+1)(x+2)(x-5)}$

Expand the numerators and combine like terms:
$\frac{x - 5 - x - 1}{(x+1)(x+2)(x-5)}$

Simplify the numerator:
$\frac{-6}{(x+1)(x+2)(x-5)}$

The simplified form of the expression is now:
$\frac{-6}{(x+1)(x+2)(x-5)}$