Answer
**Problem 28:**
The roots of the polynomial \( f(x) = -2x^4 + 14x^2 - 12x \) are:
- \( x = 0 \)
- \( x = 1 \)
- \( x = 2 \)
- \( x = -3 \)
**Problem 29:**
For the polynomial \( ax^3 + bx^2 - 2x + 8 \), the values are:
- \( a = \dfrac{19}{12} \)
- \( b = -\dfrac{41}{12} \)
Solution
### Problem 28
**Find all roots of the polynomial function \( f(x) = -2x^4 + 14x^2 - 12x \).**
**Solution:**
To find the roots, we set \( f(x) = 0 \):
\[
-2x^4 + 14x^2 - 12x = 0
\]
Factor out \( x \):
\[
x(-2x^3 + 14x - 12) = 0
\]
This gives one root:
\[
x = 0
\]
Now, solve the cubic equation:
\[
-2x^3 + 14x - 12 = 0 \quad \Rightarrow \quad 2x^3 - 14x + 12 = 0
\]
Divide the entire equation by 2:
\[
x^3 - 7x + 6 = 0
\]
**Finding Rational Roots:**
Possible rational roots are \( \pm1, \pm2, \pm3, \pm6 \).
Testing \( x = 1 \):
\[
1^3 - 7(1) + 6 = 0 \quad \Rightarrow \quad x = 1 \text{ is a root.}
\]
Factor out \( (x - 1) \):
\[
x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)
\]
Factor \( x^2 + x - 6 \):
\[
x^2 + x - 6 = (x + 3)(x - 2)
\]
**All Roots:**
\[
x = 0, \quad x = 1, \quad x = 2, \quad x = -3
\]
### Problem 29
**If \( ax^3 + bx^2 - 2x + 8 \) is divided by \( x - 2 \) and \( x + 1 \), the remainders are 3 and 5 respectively. Find the values of \( a \) and \( b \).**
**Solution:**
By the Remainder Theorem:
1. When divided by \( x - 2 \):
\[
p(2) = 3 \quad \Rightarrow \quad a(2)^3 + b(2)^2 - 2(2) + 8 = 3
\]
\[
8a + 4b - 4 + 8 = 3 \quad \Rightarrow \quad 8a + 4b + 4 = 3 \quad \Rightarrow \quad 8a + 4b = -1 \quad \text{(Equation 1)}
\]
2. When divided by \( x + 1 \):
\[
p(-1) = 5 \quad \Rightarrow \quad a(-1)^3 + b(-1)^2 - 2(-1) + 8 = 5
\]
\[
-a + b + 2 + 8 = 5 \quad \Rightarrow \quad -a + b + 10 = 5 \quad \Rightarrow \quad -a + b = -5 \quad \text{(Equation 2)}
\]
**Solving the System of Equations:**
From Equation 2:
\[
b = a - 5
\]
Substitute \( b = a - 5 \) into Equation 1:
\[
8a + 4(a - 5) = -1 \quad \Rightarrow \quad 8a + 4a - 20 = -1 \quad \Rightarrow \quad 12a = 19 \quad \Rightarrow \quad a = \frac{19}{12}
\]
Now, find \( b \):
\[
b = \frac{19}{12} - 5 = \frac{19}{12} - \frac{60}{12} = -\frac{41}{12}
\]
**Final Answer:**
- **Problem 28 Answer:** All roots are \( x = 0 \), \( x = 1 \), \( x = 2 \), and \( x = -3 \).
- **Problem 29 Answer:** \( a = \dfrac{19}{12} \) and \( b = -\dfrac{41}{12} \).
Reviewed and approved by the UpStudy tutoring team
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