14 multiple choice 1 point
factor completely.
$x^4 - x^2 - 12$
$\bigcirc\ (x + 2)(x - 2)(x^2 + 3)$
$\bigcirc\ (x^2 + 4)(x^2 - 3)$
$\bigcirc\ (x + 2)(x - 2)(x^2 - 3)$
$\bigcirc\ (x^2 - 4)(x^2 + 3)$
$\bigcirc\ (x + 4)(x - 4)(x^2 + 3)$
clear my selection
15 multiple answer 1 point
solve. $x^3 + 1 = 0$
choose all correct solutions.
$\square\ 1$
$\square\ \frac{1\pm i\sqrt{3}}{2}$
$\square\ \frac{1\pm \sqrt{5}}{2}$
$\square\ \frac{1\pm \sqrt{3}}{2}$
$\square\ -1$
16 multiple answer 1 point
which are factors of the expression? choose all that apply.
$16x^2 - 12x - 70$

Question

Accepted Answer

### Question 14 (Multiple Choice)
# Explanation:
## Step1: Let $ y = x^2 $, then the expression becomes $ y^2 - y - 12 $
## Step2: Factor $ y^2 - y - 12 $. We need two numbers that multiply to -12 and add to -1. Those numbers are -4 and 3. So, $ y^2 - y - 12=(y - 4)(y + 3) $
## Step3: Substitute back $ y = x^2 $, we get $ (x^2 - 4)(x^2 + 3) $
## Step4: Factor $ x^2 - 4 $ using difference of squares: $ x^2 - 4=(x + 2)(x - 2) $. So the completely factored form is $ (x + 2)(x - 2)(x^2 + 3) $
# Answer:
$ \boldsymbol{(x + 2)(x - 2)(x^2 + 3)} $ (the first option)

### Question 15 (Multiple Answer)
# Explanation:
## Step1: Rewrite the equation $ x^3 + 1 = 0 $ as $ x^3=-1 $ or $ x^3+1^3 = 0 $
## Step2: Use the sum of cubes formula $ a^3 + b^3=(a + b)(a^2 - ab + b^2) $. Here, $ a = x $, $ b = 1 $, so $ (x + 1)(x^2 - x + 1)=0 $
## Step3: Set each factor equal to zero:
- For $ x + 1 = 0 $, we get $ x=-1 $
- For $ x^2 - x + 1 = 0 $, use the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $ where $ a = 1 $, $ b=-1 $, $ c = 1 $. So $ x=\frac{1\pm\sqrt{1 - 4}}{2}=\frac{1\pm i\sqrt{3}}{2} $
# Answer:
$ \boldsymbol{-1} $, $ \boldsymbol{\frac{1\pm i\sqrt{3}}{2}} $ (the second and fifth options)

### Question 16 (Multiple Answer)
# Explanation:
## Step1: First, factor out the greatest common factor (GCF) of 16, -12, and -70. The GCF of 16, 12, and 70 is 2. So, $ 16x^2 - 12x - 70 = 2(8x^2 - 6x - 35) $
## Step2: Factor $ 8x^2 - 6x - 35 $. We need two numbers that multiply to $ 8\times(-35)=-280 $ and add to -6. Those numbers are -20 and 14. So, $ 8x^2 - 6x - 35=8x^2 + 14x - 20x - 35 = 2x(4x + 7)-5(4x + 7)=(2x - 5)(4x + 7) $
## Step3: So the factored form of $ 16x^2 - 12x - 70 $ is $ 2(2x - 5)(4x + 7) $. The factors are 2, $ 2x - 5 $, and $ 4x + 7 $ (assuming the options include these; since the options are not fully shown, but the process is to factor as above)
# Answer:
(Assuming the options are related to the factors 2, $ 2x - 5 $, $ 4x + 7 $, the correct factors would be these. If the options are given as, for example, 2, $ 2x - 5 $, $ 4x + 7 $, then those are the correct ones. Since the options are not fully visible, but the factoring process gives these factors.)

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Question

Explanation:

Question 14 (Multiple Choice)

Step1: Let \( y = x^2 \), then the expression becomes \( y^2 - y - 12 \)

Step2: Factor \( y^2 - y - 12 \). We need two numbers that multiply to -12 and add to -1. Those numbers are -4 and 3. So, \( y^2 - y - 12=(y - 4)(y + 3) \)

Step3: Substitute back \( y = x^2 \), we get \( (x^2 - 4)(x^2 + 3) \)

Step4: Factor \( x^2 - 4 \) using difference of squares: \( x^2 - 4=(x + 2)(x - 2) \). So the completely factored form is \( (x + 2)(x - 2)(x^2 + 3) \)

Step1: Rewrite the equation \( x^3 + 1 = 0 \) as \( x^3=-1 \) or \( x^3+1^3 = 0 \)

Step2: Use the sum of cubes formula \( a^3 + b^3=(a + b)(a^2 - ab + b^2) \). Here, \( a = x \), \( b = 1 \), so \( (x + 1)(x^2 - x + 1)=0 \)

Step3: Set each factor equal to zero:

Step1: First, factor out the greatest common factor (GCF) of 16, -12, and -70. The GCF of 16, 12, and 70 is 2. So, \( 16x^2 - 12x - 70 = 2(8x^2 - 6x - 35) \)

Step2: Factor \( 8x^2 - 6x - 35 \). We need two numbers that multiply to \( 8\times(-35)=-280 \) and add to -6. Those numbers are -20 and 14. So, \( 8x^2 - 6x - 35=8x^2 + 14x - 20x - 35 = 2x(4x + 7)-5(4x + 7)=(2x - 5)(4x + 7) \)

Step3: So the factored form of \( 16x^2 - 12x - 70 \) is \( 2(2x - 5)(4x + 7) \). The factors are 2, \( 2x - 5 \), and \( 4x + 7 \) (assuming the options include these; since the options are not fully shown, but the process is to factor as above)

Answer:

Question 15 (Multiple Answer)