sample
1. ((x^3 - 27) div (x - 3))
2. ((g^2 + 8g + 15)(g + 3)^{-1})
3. (\frac{x^3 + 3x^2 - 7x - 21}{x + 3})
4. (\frac{2x^3 + x^2 - 2x + 3}{x + 1})
5. ((x^5 + 32)(x + 2)^{-1})

Question

Accepted Answer

### Problem 1: $(x^3 - 27) \div (x - 3)$
# Explanation:
## Step 1: Recognize the formula
We know that $a^3 - b^3=(a - b)(a^2+ab + b^2)$. Here, $x^3-27=x^3 - 3^3$, so $a = x$ and $b = 3$.
So $x^3-3^3=(x - 3)(x^2+3x + 9)$
## Step 2: Divide by $(x - 3)$
Now we have $\frac{(x - 3)(x^2+3x + 9)}{x - 3}$. Since $x
eq3$ (to avoid division by zero), we can cancel out the $(x - 3)$ terms.
# Answer:
$x^2 + 3x+9$

### Problem 2: $(g^2 + 8g + 15)(g + 3)^{-1}$
# Explanation:
## Step 1: Factor the quadratic
Factor $g^2 + 8g + 15$. We need two numbers that multiply to $15$ and add to $8$. The numbers are $3$ and $5$. So $g^2+8g + 15=(g + 3)(g + 5)$
## Step 2: Simplify the expression
The expression is $\frac{(g + 3)(g + 5)}{g + 3}$. For $g
eq - 3$, we can cancel out $(g + 3)$ from numerator and denominator.
# Answer:
$g + 5$

### Problem 3: $\frac{x^3+3x^2-7x - 21}{x + 3}$
# Explanation:
## Step 1: Group the terms
Group the numerator as $(x^3+3x^2)+(-7x - 21)$
## Step 2: Factor each group
Factor $x^2$ from the first group: $x^2(x + 3)$
Factor $-7$ from the second group: $-7(x + 3)$
So the numerator becomes $x^2(x + 3)-7(x + 3)=(x + 3)(x^2-7)$
## Step 3: Divide by $(x + 3)$
Now we have $\frac{(x + 3)(x^2-7)}{x + 3}$. For $x
eq - 3$, we can cancel out $(x + 3)$
# Answer:
$x^2-7$

### Problem 4: $\frac{2x^3+x^2-2x + 3}{x + 1}$
# Explanation:
## Step 1: Use polynomial long division
Divide $2x^3+x^2-2x + 3$ by $x + 1$
- First term: $\frac{2x^3}{x}=2x^2$. Multiply $(x + 1)$ by $2x^2$ to get $2x^3+2x^2$. Subtract from the dividend: $(2x^3+x^2-2x + 3)-(2x^3+2x^2)=-x^2-2x + 3$
- Second term: $\frac{-x^2}{x}=-x$. Multiply $(x + 1)$ by $-x$ to get $-x^2-x$. Subtract: $(-x^2-2x + 3)-(-x^2-x)=-x + 3$
- Third term: $\frac{-x}{x}=-1$. Multiply $(x + 1)$ by $-1$ to get $-x - 1$. Subtract: $(-x + 3)-(-x - 1)=4$
So the division gives $2x^2-x - 1+\frac{4}{x + 1}$ (or if we consider only the quotient when there is a remainder, but maybe we made a mistake in calculation? Wait, let's check again. Wait, maybe synthetic division is better.
Using synthetic division with root $x=-1$
Coefficients: $2$, $1$, $-2$, $3$
Bring down $2$
Multiply by $-1$: $2\times(-1)=-2$. Add to next coefficient: $1+(-2)=-1$
Multiply by $-1$: $-1\times(-1)=1$. Add to next coefficient: $-2 + 1=-1$
Multiply by $-1$: $-1\times(-1)=1$. Add to last coefficient: $3+1 = 4$
So the quotient is $2x^2-x - 1$ and remainder $4$. So $\frac{2x^3+x^2-2x + 3}{x + 1}=2x^2-x - 1+\frac{4}{x + 1}$
But maybe there was a miscalculation in the problem? Wait, let's check the original problem again. The numerator is $2x^3+x^2-2x + 3$? Maybe a typo? If the numerator was $2x^3+x^2-2x - 3$, then the remainder would be zero. Assuming the problem is correct as given, the result is $2x^2-x - 1+\frac{4}{x + 1}$
# Answer:
$2x^2 - x - 1+\frac{4}{x + 1}$ (or if we consider the problem might have a typo, but based on given, this is the result)

### Problem 5: $(x^5 + 32)(x + 2)^{-1}$
# Explanation:
## Step 1: Recognize the formula
We know that $a^5 + b^5=(a + b)(a^4-a^3b+a^2b^2-ab^3+b^4)$. Here, $x^5+32=x^5+2^5$, so $a = x$ and $b = 2$
So $x^5+2^5=(x + 2)(x^4-2x^3+4x^2-8x + 16)$
## Step 2: Simplify the expression
The expression is $\frac{(x + 2)(x^4-2x^3+4x^2-8x + 16)}{x + 2}$. For $x
eq - 2$, we can cancel out $(x + 2)$
# Answer:
$x^4-2x^3+4x^2-8x + 16$

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QUESTION IMAGE

Question

Explanation:

Problem 1: \((x^3 - 27) \div (x - 3)\)

Step 1: Recognize the formula

Step 2: Divide by \((x - 3)\)

Step 1: Factor the quadratic

Step 2: Simplify the expression

Step 1: Group the terms

Step 2: Factor each group

Step 3: Divide by \((x + 3)\)

Answer:

Problem 2: \((g^2 + 8g + 15)(g + 3)^{-1}\)