4-4 practice
adding and subtracting rational expressions
find the lcm for each group of expressions.
1. $2x^2 - 8x + 8$ and $3x^2 + 27x - 30$
2. $4x^2 + 12x + 9$ and $4x^2 - 9$
3. $2x^2 - 18$ and $5x^3 + 30x^2 + 45x$
find the sum.
4. $\frac{6y - 4}{y^2 - 5} + \frac{3y + 1}{y^2 - 5}$
5. $\frac{x + 2}{x^2 + 4x + 4} + \frac{2}{x + 2}$
6. $\frac{4}{y^2 - 25} + \frac{6}{x^2 + 6x + 5}$

Question

Accepted Answer

### Problem 1: Find the LCM of $2x^2 - 8x + 8$ and $3x^2 + 27x - 30$
# Explanation:
## Step 1: Factor each expression
First, factor $2x^2 - 8x + 8$:
$$
\begin{align*}
2x^2 - 8x + 8&=2(x^2 - 4x + 4)\
&=2(x - 2)^2
\end{align*}
$$
Then, factor $3x^2 + 27x - 30$:
$$
\begin{align*}
3x^2 + 27x - 30&=3(x^2 + 9x - 10)\
&=3(x + 10)(x - 1)
\end{align*}
$$
## Step 2: Identify the LCM
The LCM is the product of the highest power of each factor that appears in the factorizations.
The factors are $2$, $3$, $(x - 2)^2$, $(x + 10)$, and $(x - 1)$.
So, the LCM is $2\times3\times(x - 2)^2\times(x + 10)\times(x - 1)=6(x - 2)^2(x + 10)(x - 1)$
# Answer:
$6(x - 2)^2(x + 10)(x - 1)$

### Problem 2: Find the LCM of $4x^2 + 12x + 9$ and $4x^2 - 9$
# Explanation:
## Step 1: Factor each expression
Factor $4x^2 + 12x + 9$:
$$
4x^2 + 12x + 9=(2x + 3)^2
$$
Factor $4x^2 - 9$ (difference of squares):
$$
4x^2 - 9=(2x + 3)(2x - 3)
$$
## Step 2: Identify the LCM
The highest power of $(2x + 3)$ is $(2x + 3)^2$, and we also have $(2x - 3)$.
So, the LCM is $(2x + 3)^2(2x - 3)$
# Answer:
$(2x + 3)^2(2x - 3)$

### Problem 3: Find the LCM of $2x^2 - 18$ and $5x^3 + 30x^2 + 45x$
# Explanation:
## Step 1: Factor each expression
Factor $2x^2 - 18$:
$$
\begin{align*}
2x^2 - 18&=2(x^2 - 9)\
&=2(x + 3)(x - 3)
\end{align*}
$$
Factor $5x^3 + 30x^2 + 45x$:
$$
\begin{align*}
5x^3 + 30x^2 + 45x&=5x(x^2 + 6x + 9)\
&=5x(x + 3)^2
\end{align*}
$$
## Step 2: Identify the LCM
The factors are $2$, $5$, $x$, $(x + 3)^2$, and $(x - 3)$.
So, the LCM is $2\times5\times x\times(x + 3)^2\times(x - 3)=10x(x + 3)^2(x - 3)$
# Answer:
$10x(x + 3)^2(x - 3)$

### Problem 4: Find the sum $\frac{6y - 4}{y^2 - 5}+\frac{3y + 1}{y^2 - 5}$
# Explanation:
## Step 1: Add the numerators (same denominator)
Since the denominators are the same ($y^2 - 5$), we add the numerators:
$$
\frac{(6y - 4)+(3y + 1)}{y^2 - 5}
$$
## Step 2: Simplify the numerator
$$
\begin{align*}
(6y - 4)+(3y + 1)&=6y - 4 + 3y + 1\
&=9y - 3
\end{align*}
$$
So, the expression becomes $\frac{9y - 3}{y^2 - 5}$. We can factor the numerator: $9y - 3 = 3(3y - 1)$, so $\frac{3(3y - 1)}{y^2 - 5}$
# Answer:
$\frac{3(3y - 1)}{y^2 - 5}$ (or $\frac{9y - 3}{y^2 - 5}$)

### Problem 5: Find the sum $\frac{x + 2}{x^2 + 4x + 4}+\frac{2}{x + 2}$
# Explanation:
## Step 1: Factor the denominator of the first fraction
Factor $x^2 + 4x + 4=(x + 2)^2$. So the first fraction is $\frac{x + 2}{(x + 2)^2}=\frac{1}{x + 2}$ (for $x
eq - 2$)
## Step 2: Add the fractions
Now we have $\frac{1}{x + 2}+\frac{2}{x + 2}$. Since the denominators are the same, add the numerators:
$$
\frac{1 + 2}{x + 2}=\frac{3}{x + 2}
$$
# Answer:
$\frac{3}{x + 2}$

### Problem 6: Find the sum $\frac{4}{y^2 - 25}+\frac{6}{x^2 + 6x + 5}$
# Explanation:
## Step 1: Factor each denominator
Factor $y^2 - 25=(y + 5)(y - 5)$ (difference of squares).
Factor $x^2 + 6x + 5=(x + 1)(x + 5)$ (by finding two numbers that multiply to 5 and add to 6: 1 and 5).
## Step 2: Find the LCM of the denominators
The denominators are $(y + 5)(y - 5)$ and $(x + 1)(x + 5)$. The LCM is $(y + 5)(y - 5)(x + 1)(x + 5)$
## Step 3: Rewrite each fraction with the LCM as the denominator
For the first fraction: $\frac{4}{(y + 5)(y - 5)}=\frac{4(x + 1)(x + 5)}{(y + 5)(y - 5)(x + 1)(x + 5)}$
For the second fraction: $\frac{6}{(x + 1)(x + 5)}=\frac{6(y + 5)(y - 5)}{(y + 5)(y - 5)(x + 1)(x + 5)}$
## Step 4: Add the numerators
$$
\begin{align*}
&\frac{4(x + 1)(x + 5)+6(y + 5)(y - 5)}{(y + 5)(y - 5)(x + 1)(x + 5)}\
&=\frac{4(x^2 + 6x + 5)+6(y^2 - 25)}{(y + 5)(y - 5)(x + 1)(x + 5)}\
&=\frac{4x^2 + 24x + 20 + 6y^2 - 150}{(y + 5)(y - 5)(x + 1)(x + 5)}\
&=\frac{4x^2 + 24x + 6y^2 - 130}{(y + 5)(y - 5)(x + 1)(x + 5)}
\end{align*}
$$
We can factor out a 2 from the numerator: $\frac{2(2x^2 + 12x + 3y^2 - 65)}{(y + 5)(y - 5)(x + 1)(x + 5)}$
# Answer:
$\frac{2(2x^2 + 12x + 3y^2 - 65)}{(y + 5)(y - 5)(x + 1)(x + 5)}$ (or the expanded numerator form)

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QUESTION IMAGE

Question

Explanation:

Problem 1: Find the LCM of \(2x^2 - 8x + 8\) and \(3x^2 + 27x - 30\)

Step 1: Factor each expression

Step 2: Identify the LCM

Step 1: Factor each expression

Step 2: Identify the LCM

Step 1: Factor each expression

Step 2: Identify the LCM

Answer:

Problem 2: Find the LCM of \(4x^2 + 12x + 9\) and \(4x^2 - 9\)