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12) $y = x^4 + 7x^2 - 8$
13) $y = x^5 - 15x^4 + 27x^3 - 13x^2$
14) $y = x^2 - 12x - 13$
15) $y = x^3 - 14x + 20$

Question

Accepted Answer

### Problem 12: $ y = x^4 + 7x^2 - 8 $
# Explanation:
## Step 1: Substitute $ u = x^2 $
Let $ u = x^2 $, then the equation becomes $ y = u^2 + 7u - 8 $.
## Step 2: Factor the quadratic
Factor $ u^2 + 7u - 8 $: $ (u + 8)(u - 1) = 0 $.
## Step 3: Solve for $ u $
Set each factor to zero: $ u + 8 = 0 $ or $ u - 1 = 0 $, so $ u = -8 $ or $ u = 1 $.
## Step 4: Substitute back $ u = x^2 $
- For $ u = -8 $: $ x^2 = -8 $, so $ x = \pm 2i\sqrt{2} $.
- For $ u = 1 $: $ x^2 = 1 $, so $ x = \pm 1 $.

# Answer:
The zeros are $ x = 1, -1, 2i\sqrt{2}, -2i\sqrt{2} $.

### Problem 13: $ y = x^5 - 15x^4 + 27x^3 - 13x^2 $
# Explanation:
## Step 1: Factor out $ x^2 $
Factor $ x^2 $ from the polynomial: $ y = x^2(x^3 - 15x^2 + 27x - 13) $.
## Step 2: Find a root of the cubic
Test $ x = 1 $ in $ x^3 - 15x^2 + 27x - 13 $: $ 1 - 15 + 27 - 13 = 0 $, so $ (x - 1) $ is a factor.
## Step 3: Perform polynomial division or use synthetic division
Divide $ x^3 - 15x^2 + 27x - 13 $ by $ (x - 1) $. Using synthetic division:
$$
\begin{array}{r|rrrr}
1 & 1 & -15 & 27 & -13 \
  &   & 1  & -14 & 13 \
\hline
  & 1 & -14 & 13 & 0 \
\end{array}
$$
So, $ x^3 - 15x^2 + 27x - 13 = (x - 1)(x^2 - 14x + 13) $.
## Step 4: Factor the quadratic
Factor $ x^2 - 14x + 13 $: $ (x - 1)(x - 13) = 0 $.
## Step 5: Combine all factors
The polynomial becomes $ y = x^2(x - 1)(x - 1)(x - 13) = x^2(x - 1)^2(x - 13) $.
## Step 6: Find zeros
Set each factor to zero: $ x^2 = 0 $ (double root $ x = 0 $), $ (x - 1)^2 = 0 $ (double root $ x = 1 $), and $ x - 13 = 0 $ (root $ x = 13 $).

# Answer:
The zeros are $ x = 0 $ (multiplicity 2), $ x = 1 $ (multiplicity 2), and $ x = 13 $.

### Problem 14: $ y = x^2 - 12x - 13 $
# Explanation:
## Step 1: Factor the quadratic
Factor $ x^2 - 12x - 13 $: Find two numbers that multiply to -13 and add to -12. The numbers are -13 and 1. So, $ (x - 13)(x + 1) = 0 $.
## Step 2: Solve for $ x $
Set each factor to zero: $ x - 13 = 0 $ or $ x + 1 = 0 $, so $ x = 13 $ or $ x = -1 $.

# Answer:
The zeros are $ x = 13 $ and $ x = -1 $.

### Problem 15: $ y = x^3 - 14x + 20 $
# Explanation:
## Step 1: Test possible rational roots
Using Rational Root Theorem, possible roots are $ \pm1, \pm2, \pm4, \pm5, \pm10, \pm20 $. Test $ x = 2 $: $ 8 - 28 + 20 = 0 $, so $ (x - 2) $ is a factor.
## Step 2: Perform polynomial division or use synthetic division
Divide $ x^3 - 14x + 20 $ by $ (x - 2) $. Using synthetic division:
$$
\begin{array}{r|rrrr}
2 & 1 & 0 & -14 & 20 \
  &   & 2 & 4 & -20 \
\hline
  & 1 & 2 & -10 & 0 \
\end{array}
$$
So, the polynomial becomes $ (x - 2)(x^2 + 2x - 10) $.
## Step 3: Solve the quadratic
For $ x^2 + 2x - 10 = 0 $, use quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ with $ a = 1 $, $ b = 2 $, $ c = -10 $:
$ x = \frac{-2 \pm \sqrt{4 + 40}}{2} = \frac{-2 \pm \sqrt{44}}{2} = \frac{-2 \pm 2\sqrt{11}}{2} = -1 \pm \sqrt{11} $.

# Answer:
The zeros are $ x = 2 $, $ x = -1 + \sqrt{11} $, and $ x = -1 - \sqrt{11} $.

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QUESTION IMAGE

Question

Explanation:

Problem 12: \( y = x^4 + 7x^2 - 8 \)

Step 1: Substitute \( u = x^2 \)

Step 2: Factor the quadratic

Step 3: Solve for \( u \)

Step 4: Substitute back \( u = x^2 \)

Step 1: Factor out \( x^2 \)

Step 2: Find a root of the cubic

Step 3: Perform polynomial division or use synthetic division

Step 4: Factor the quadratic

Step 5: Combine all factors

Step 6: Find zeros

Step 1: Factor the quadratic

Step 2: Solve for \( x \)

Answer:

Problem 13: \( y = x^5 - 15x^4 + 27x^3 - 13x^2 \)