check in 5-2 factoring polynomial functions
identify the zeros, their multiplicity, and their effect on the graph of the function.
1. $f(x) = -x(3x - 2)^3(x + 9)^5$
2. $f(x) = x^3 + 10x^2 + 25x$
solve each equation by factoring. simplify all irrational and complex solutions.
3. $2x^4 = 50$
4. $x^3 + 216 = 0$
5. $x^3 = x^2 + 20x$
6. $x^4 + 13x^2 + 40 = 0$
7. $x^4 - x^2 = x^2 + 8$
8. $4x^3 + 5x^2 - 16x - 20 = 0$

Question

Accepted Answer

### Problem 1: $ f(x) = -x(3x - 2)^3(x + 9)^5 $
# Explanation:
## Step 1: Find Zeros
To find the zeros, set each factor equal to zero:
- For $ -x = 0 $, we get $ x = 0 $.
- For $ (3x - 2)^3 = 0 $, solve $ 3x - 2 = 0 $, so $ x = \frac{2}{3} $.
- For $ (x + 9)^5 = 0 $, solve $ x + 9 = 0 $, so $ x = -9 $.

## Step 2: Determine Multiplicity
The multiplicity of a zero is the exponent of its corresponding factor:
- For $ x = 0 $, the factor is $ -x $ (exponent 1), so multiplicity is 1.
- For $ x = \frac{2}{3} $, the factor is $ (3x - 2)^3 $ (exponent 3), so multiplicity is 3.
- For $ x = -9 $, the factor is $ (x + 9)^5 $ (exponent 5), so multiplicity is 5.

## Step 3: Determine Effect on Graph
- If multiplicity is odd, the graph crosses the x - axis at that zero.
- If multiplicity is even, the graph touches the x - axis and turns around at that zero.
- For $ x = 0 $ (multiplicity 1, odd): The graph crosses the x - axis at $ x = 0 $.
- For $ x = \frac{2}{3} $ (multiplicity 3, odd): The graph crosses the x - axis at $ x=\frac{2}{3} $.
- For $ x = -9 $ (multiplicity 5, odd): The graph crosses the x - axis at $ x = -9 $.

# Answer:
| Zero | Multiplicity | Effect |
|------|--------------|--------|
| $ 0 $ | $ 1 $ | Crosses the x - axis |
| $ \frac{2}{3} $ | $ 3 $ | Crosses the x - axis |
| $ -9 $ | $ 5 $ | Crosses the x - axis |

### Problem 2: $ f(x)=x^{3}+10x^{2}+25x $
# Explanation:
## Step 1: Factor the Polynomial
First, factor out the greatest common factor $ x $: $ f(x)=x(x^{2}+10x + 25) $.
Then, factor the quadratic $ x^{2}+10x + 25 $. We know that $ x^{2}+10x + 25=(x + 5)^{2} $ (since $ (a + b)^2=a^{2}+2ab + b^{2} $, here $ a=x $, $ b = 5 $, $ 2ab=10x $, $ a^{2}=x^{2} $, $ b^{2}=25 $). So $ f(x)=x(x + 5)^{2} $.

## Step 2: Find Zeros
Set each factor equal to zero:
- For $ x = 0 $, the factor is $ x $, so $ x = 0 $.
- For $ (x + 5)^{2}=0 $, solve $ x+5 = 0 $, so $ x=-5 $.

## Step 3: Determine Multiplicity
- For $ x = 0 $, the factor is $ x $ (exponent 1), so multiplicity is 1.
- For $ x=-5 $, the factor is $ (x + 5)^{2} $ (exponent 2), so multiplicity is 2.

## Step 4: Determine Effect on Graph
- For $ x = 0 $ (multiplicity 1, odd): The graph crosses the x - axis at $ x = 0 $.
- For $ x=-5 $ (multiplicity 2, even): The graph touches the x - axis and turns around at $ x=-5 $.

# Answer:
| Zero | Multiplicity | Effect |
|------|--------------|--------|
| $ 0 $ | $ 1 $ | Crosses the x - axis |
| $ -5 $ | $ 2 $ | Touches the x - axis and turns around |

### Problem 3: $ 2x^{4}=50 $
# Explanation:
## Step 1: Isolate $ x^{4} $
Divide both sides of the equation by 2:
$ \frac{2x^{4}}{2}=\frac{50}{2} $
$ x^{4}=25 $

## Step 2: Solve for $ x $
We can rewrite $ x^{4}=25 $ as $ x^{4}-25 = 0 $. Notice that $ x^{4}-25=(x^{2})^{2}-5^{2} $, and using the difference of squares formula $ a^{2}-b^{2}=(a + b)(a - b) $, we have $ (x^{2}+5)(x^{2}-5)=0 $.
- For $ x^{2}+5 = 0 $, we get $ x^{2}=-5 $, so $ x=\pm\sqrt{-5}=\pm i\sqrt{5} $ (where $ i=\sqrt{-1} $).
- For $ x^{2}-5 = 0 $, we get $ x^{2}=5 $, so $ x=\pm\sqrt{5} $.

# Answer:
$ x=\sqrt{5},x = -\sqrt{5},x=i\sqrt{5},x=-i\sqrt{5} $

### Problem 4: $ x^{3}+216 = 0 $
# Explanation:
## Step 1: Rewrite as Sum of Cubes
We know that $ a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2}) $. Here, $ x^{3}+216=x^{3}+6^{3} $, so $ (x + 6)(x^{2}-6x + 36)=0 $.

## Step 2: Solve for $ x $
- For $ x + 6=0 $, we get $ x=-6 $.
- For $ x^{2}-6x + 36 = 0 $, use the quadratic formula $ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} $, where $ a = 1 $, $ b=-6 $, $ c = 36 $.
$ \Delta=b^{2}-4ac=(-6)^{2}-4\times1\times36=36 - 144=-108 $
$ x=\frac{6\pm\sqrt{-108}}{2}=\frac{6\pm6i\sqrt{3}}{2}=3\pm3i\sqrt{3} $

# Answer:
$ x=-6,x = 3 + 3i\sqrt{3},x=3 - 3i\sqrt{3} $

### Problem 5: $ x^{3}=x^{2}+20x $
# Explanation:
## Step 1: Move All Terms to One Side
Subtract $ x^{2} $ and $ 20x $ from both sides:
$ x^{3}-x^{2}-20x = 0 $

## Step 2: Factor Out $ x $
$ x(x^{2}-x - 20)=0 $

## Step 3: Factor the Quadratic
Factor $ x^{2}-x - 20 $. We need two numbers that multiply to - 20 and add to - 1. The numbers are - 5 and 4. So $ x^{2}-x - 20=(x - 5)(x + 4) $.
The equation becomes $ x(x - 5)(x + 4)=0 $.

## Step 4: Solve for $ x $
Set each factor equal to zero:
- $ x = 0 $
- $ x - 5=0\Rightarrow x = 5 $
- $ x + 4=0\Rightarrow x=-4 $

# Answer:
$ x = 0,x = 5,x=-4 $

### Problem 6: $ x^{4}+13x^{2}+40 = 0 $
# Explanation:
## Step 1: Let $ u=x^{2} $
Substitute $ u $ into the equation: $ u^{2}+13u + 40 = 0 $

## Step 2: Factor the Quadratic in $ u $
We need two numbers that multiply to 40 and add to 13. The numbers are 5 and 8. So $ (u + 5)(u + 8)=0 $.

## Step 3: Substitute Back $ u=x^{2} $
- For $ u+5 = 0\Rightarrow x^{2}+5 = 0\Rightarrow x^{2}=-5\Rightarrow x=\pm i\sqrt{5} $
- For $ u + 8=0\Rightarrow x^{2}+8 = 0\Rightarrow x^{2}=-8\Rightarrow x=\pm i\sqrt{8}=\pm2i\sqrt{2} $

# Answer:
$ x=i\sqrt{5},x=-i\sqrt{5},x = 2i\sqrt{2},x=-2i\sqrt{2} $

### Problem 7: $ x^{4}-x^{2}=x^{2}+8 $
# Explanation:
## Step 1: Move All Terms to One Side
Subtract $ x^{2} $ and 8 from both sides:
$ x^{4}-2x^{2}-8 = 0 $

## Step 2: Let $ u=x^{2} $
Substitute $ u $ into the equation: $ u^{2}-2u - 8 = 0 $

## Step 3: Factor the Quadratic in $ u $
We need two numbers that multiply to - 8 and add to - 2. The numbers are - 4 and 2. So $ (u - 4)(u + 2)=0 $.

## Step 4: Substitute Back $ u=x^{2} $
- For $ u - 4=0\Rightarrow x^{2}-4 = 0\Rightarrow(x - 2)(x + 2)=0\Rightarrow x = 2,x=-2 $
- For $ u + 2=0\Rightarrow x^{2}+2 = 0\Rightarrow x^{2}=-2\Rightarrow x=\pm i\sqrt{2} $

# Answer:
$ x = 2,x=-2,x=i\sqrt{2},x=-i\sqrt{2} $

### Problem 8: $ 4x^{3}+5x^{2}-16x - 20 = 0 $
# Explanation:
## Step 1: Group Terms
Group the first two terms and the last two terms:
$ (4x^{3}+5x^{2})+(-16x - 20)=0 $

## Step 2: Factor Out Common Factors from Each Group
- From $ 4x^{3}+5x^{2} $, factor out $ x^{2} $: $ x^{2}(4x + 5) $
- From $ -16x - 20 $, factor out - 4: $ -4(4x + 5) $
The equation becomes $ x^{2}(4x + 5)-4(4x + 5)=0 $

## Step 3: Factor Out $ (4x + 5) $
$ (4x + 5)(x^{2}-4)=0 $

## Step 4: Factor $ x^{2}-4 $
Using the difference of squares formula, $ x^{2}-4=(x - 2)(x + 2) $. So the equation is $ (4x + 5)(x - 2)(x + 2)=0 $

## Step 5: Solve for $ x $
- $ 4x+5 = 0\Rightarrow x=-\frac{5}{4} $
- $ x - 2=0\Rightarrow x = 2 $
- $ x + 2=0\Rightarrow x=-2 $

# Answer:
$ x=-\frac{5}{4},x = 2,x=-2 $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: \( f(x) = -x(3x - 2)^3(x + 9)^5 \)

Step 1: Find Zeros

Step 2: Determine Multiplicity

Step 3: Determine Effect on Graph

Step 1: Factor the Polynomial

Step 2: Find Zeros

Step 3: Determine Multiplicity

Step 4: Determine Effect on Graph

Step 1: Isolate \( x^{4} \)

Step 2: Solve for \( x \)

Answer:

Problem 2: \( f(x)=x^{3}+10x^{2}+25x \)

Zero	Multiplicity	Effect
\( \frac{2}{3} \)	\( 3 \)	Crosses the x - axis
\( -9 \)	\( 5 \)	Crosses the x - axis