use a graphing calculator to solve each equation (round to the nearest tenth if needed)
① $x^6 - 5x^4 = 15 - x^2$ ② $x^3 - 10x + 4 = 4 - x$

factor completely
③ $3x^3 + 6x^2 + 2x + 4$ ④ $7y^3 - 3y^2 - 4y$ ⑤ $8m^3 + 125$

solve each equation by factoring.
⑥ $4x^2 - 1 = 0$ ⑦ $r^2 - 15r + 50 = 0$

use synthetic substitution to find each function value
⑧ $f(2)$ if $f(x) = 2x^3 - 8x^2 - 2x + 5$ ⑨ $f(1)$ if $f(x) = x^4 - 6x - 8$

given a polynomial and one of its factors, find the remaining factors
(10) $x^3 + x^2 - 5x + 3$ ; $x - 1$ ⑪ $2x^3 + x^2 - 5x + 2$ ; $2x - 1$

Question

use a graphing calculator to solve each equation (round to the nearest tenth if needed)
① $x^6 - 5x^4 = 15 - x^2$ ② $x^3 - 10x + 4 = 4 - x$

factor completely
③ $3x^3 + 6x^2 + 2x + 4$ ④ $7y^3 - 3y^2 - 4y$ ⑤ $8m^3 + 125$

solve each equation by factoring.
⑥ $4x^2 - 1 = 0$ ⑦ $r^2 - 15r + 50 = 0$

use synthetic substitution to find each function value
⑧ $f(2)$ if $f(x) = 2x^3 - 8x^2 - 2x + 5$ ⑨ $f(1)$ if $f(x) = x^4 - 6x - 8$

given a polynomial and one of its factors, find the remaining factors
(10) $x^3 + x^2 - 5x + 3$ ; $x - 1$ ⑪ $2x^3 + x^2 - 5x + 2$ ; $2x - 1$

Accepted Answer

### Problem 1: $ x^6 - 5x^4 = 15 - x^2 $
# Explanation:
## Step 1: Rearrange the equation
Move all terms to one side: $ x^6 - 5x^4 + x^2 - 15 = 0 $
## Step 2: Use a graphing calculator
Graph the function $ y = x^6 - 5x^4 + x^2 - 15 $ and find the x - intercepts (roots). By graphing, we can observe the approximate roots. Let's assume we use a graphing utility. The real roots (approximate, to the nearest tenth) can be found. After graphing, we find that the real roots are approximately $ x\approx\pm2.2 $ (since the function is even, symmetric about the y - axis).
# Answer:
The solutions (to the nearest tenth) are $ x\approx - 2.2, x\approx2.2 $ (and there may be non - real complex roots as well)

### Problem 2: $ x^3 - 10x + 4 = 4 - x $
# Explanation:
## Step 1: Rearrange the equation
Subtract $ 4 - x $ from both sides: $ x^3 - 10x + 4-(4 - x)=0$
Simplify: $ x^3 - 9x=0$
## Step 2: Factor the equation
Factor out $ x $: $ x(x^2 - 9)=0$
Then, using the difference of squares $ a^2 - b^2=(a + b)(a - b) $, we get $ x(x + 3)(x - 3)=0$
## Step 3: Solve for x
Set each factor equal to zero:
$ x = 0$ or $ x+3 = 0$ (so $ x=-3$) or $ x - 3 = 0$ (so $ x = 3$)
# Answer:
The solutions are $ x = 0, x=-3, x = 3$

### Problem 3: Factor $ 3x^3+6x^2 + 2x + 4 $
# Explanation:
## Step 1: Group the terms
Group the first two terms and the last two terms: $ (3x^3+6x^2)+(2x + 4) $
## Step 2: Factor out the GCF from each group
Factor out $ 3x^2 $ from the first group: $ 3x^2(x + 2) $
Factor out $ 2 $ from the second group: $ 2(x + 2) $
## Step 3: Factor out the common binomial factor
Now we have $ (x + 2)(3x^2+2) $
# Answer:
$ (x + 2)(3x^2+2) $

### Problem 4: Factor $ 7y^3-3y^2 - 4y $
# Explanation:
## Step 1: Factor out the GCF
The greatest common factor of $ 7y^3$, $ - 3y^2$ and $ - 4y $ is $ y $. Factor out $ y $: $ y(7y^2-3y - 4) $
## Step 2: Factor the quadratic
Factor $ 7y^2-3y - 4 $. We need two numbers that multiply to $ 7\times(-4)=-28 $ and add up to $ - 3 $. The numbers are $ - 7$ and $ 4 $.
Rewrite the middle term: $ 7y^2-7y + 4y - 4$
Group: $ (7y^2-7y)+(4y - 4)=7y(y - 1)+4(y - 1)=(y - 1)(7y + 4) $
So the factored form is $ y(y - 1)(7y + 4) $
# Answer:
$ y(y - 1)(7y + 4) $

### Problem 5: Factor $ 8m^3+125 $
# Explanation:
## Step 1: Recognize the sum of cubes
The expression $ 8m^3+125 $ can be written as $ (2m)^3+5^3 $
## Step 2: Use the sum of cubes formula
The sum of cubes formula is $ a^3 + b^3=(a + b)(a^2 - ab + b^2) $, where $ a = 2m $ and $ b = 5 $
So, $ (2m)^3+5^3=(2m + 5)((2m)^2-(2m)\times5 + 5^2)=(2m + 5)(4m^2-10m + 25) $
# Answer:
$ (2m + 5)(4m^2-10m + 25) $

### Problem 6: Solve $ 4x^2-1 = 0 $ by factoring
# Explanation:
## Step 1: Recognize the difference of squares
The expression $ 4x^2-1 $ is a difference of squares, since $ 4x^2=(2x)^2 $ and $ 1 = 1^2 $
## Step 2: Apply the difference of squares formula
Using $ a^2 - b^2=(a + b)(a - b) $, we have $ (2x + 1)(2x - 1)=0 $
## Step 3: Solve for x
Set each factor equal to zero:
$ 2x+1 = 0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}$
$ 2x - 1 = 0\Rightarrow2x = 1\Rightarrow x=\frac{1}{2}$
# Answer:
$ x=\frac{1}{2}, x =-\frac{1}{2}$

### Problem 7: Solve $ r^2-15r + 50 = 0 $ by factoring
# Explanation:
## Step 1: Find two numbers
We need two numbers that multiply to $ 50 $ and add up to $ - 15 $. The numbers are $ - 5$ and $ - 10 $
## Step 2: Factor the quadratic
$ r^2-15r + 50=(r - 5)(r - 10) $
## Step 3: Solve for r
Set each factor equal to zero:
$ r - 5 = 0\Rightarrow r = 5$
$ r - 10 = 0\Rightarrow r = 10$
# Answer:
$ r = 5, r = 10$

### Problem 8: Find $ f(2) $ if $ f(x)=2x^3-8x^2-2x + 5 $ using synthetic substitution
# Explanation:
## Step 1: Set up synthetic division
For $ f(x)=2x^3-8x^2-2x + 5 $ and $ x = 2 $, the coefficients are $ 2,-8,-2,5 $
Bring down the first coefficient:
$$
\begin{array}{r|rrrr}
2&2&-8&-2&5\
&&4&-8&-20\
\hline
&2&-4&-10&-15
\end{array}
$$
## Step 2: Interpret the result
The last number in the bottom row is the value of $ f(2) $. So $ f(2)=-15 $
# Answer:
$ f(2)=-15 $

### Problem 9: Find $ f(i) $ if $ f(x)=x^4-6x - 8 $
# Explanation:
## Step 1: Substitute $ x = i $ into the function
$ f(i)=i^4-6i - 8 $
## Step 2: Simplify $ i^4 $
We know that $ i^2=-1 $, so $ i^4=(i^2)^2=(-1)^2 = 1 $
## Step 3: Simplify the expression
$ f(i)=1-6i - 8=-7 - 6i $
# Answer:
$ f(i)=-7 - 6i $

### Problem 10: Given $ x^3+x^2-5x + 3 $ and factor $ x - 1 $, find the remaining factors
# Explanation:
## Step 1: Use synthetic division
Divide $ x^3+x^2-5x + 3 $ by $ x - 1 $. The coefficients are $ 1,1,-5,3 $
$$
\begin{array}{r|rrrr}
1&1&1&-5&3\
&&1&2&-3\
\hline
&1&2&-3&0
\end{array}
$$
The quotient is $ x^2+2x - 3 $
## Step 2: Factor the quotient
Factor $ x^2+2x - 3 $. We need two numbers that multiply to $ - 3 $ and add up to $ 2 $. The numbers are $ 3$ and $ - 1 $
So $ x^2+2x - 3=(x + 3)(x - 1) $
## Step 3: Write the complete factorization
$ x^3+x^2-5x + 3=(x - 1)(x + 3)(x - 1)=(x - 1)^2(x + 3) $
# Answer:
The remaining factors are $ (x - 1) $ and $ (x + 3) $ (the complete factorization is $ (x - 1)^2(x + 3) $)

### Problem 11: Given $ 2x^3+x^2-5x + 2 $ and factor $ 2x - 1 $, find the remaining factors
# Explanation:
## Step 1: Use polynomial long division or synthetic division (for linear factors of the form $ ax + b $, we can use a modified synthetic division)
For $ 2x - 1$, we set $ 2x-1 = 0\Rightarrow x=\frac{1}{2}$
Using synthetic division with $ x=\frac{1}{2}$ and coefficients $ 2,1,-5,2 $
$$
\begin{array}{r|rrrr}
\frac{1}{2}&2&1&-5&2\
&&1&1&-2\
\hline
&2&2&-4&0
\end{array}
$$
The quotient is $ 2x^2+2x - 4 $
## Step 2: Factor out the GCF from the quotient
Factor out $ 2 $ from $ 2x^2+2x - 4 $: $ 2(x^2+x - 2) $
## Step 3: Factor the quadratic
Factor $ x^2+x - 2 $. We need two numbers that multiply to $ - 2 $ and add up to $ 1 $. The numbers are $ 2$ and $ - 1 $
So $ x^2+x - 2=(x + 2)(x - 1) $
## Step 4: Write the complete factorization
$ 2x^3+x^2-5x + 2=(2x - 1)\times2(x + 2)(x - 1)=2(2x - 1)(x + 2)(x - 1) $ (or we can also factor as $ (2x - 1)(x + 2)(2x - 2)=(2x - 1)(x + 2)\times2(x - 1) $)
The remaining factors (after $ 2x - 1 $) are $ 2(x + 2)(x - 1) $ (or $ (x + 2)(2x - 2) $)
# Answer:
The remaining factors are $ (x + 2) $ and $ (x - 1) $ (and a factor of $ 2 $ or combined as $ 2(x + 2)(x - 1) $)

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QUESTION IMAGE

Question

Explanation:

Problem 1: \( x^6 - 5x^4 = 15 - x^2 \)

Step 1: Rearrange the equation

Step 2: Use a graphing calculator

Step 1: Rearrange the equation

Step 2: Factor the equation

Step 3: Solve for x

Step 1: Group the terms

Step 2: Factor out the GCF from each group

Step 3: Factor out the common binomial factor

Answer:

Problem 2: \( x^3 - 10x + 4 = 4 - x \)