Solving Polynomial Equations Worksheets with Answer Key - Free Printable

Problem: Solving Polynomial Equations

We are tasked with solving the given polynomial equations. Below is a detailed solution for each equation.

---

#### Equation 1: \( 6a^3 - 42a = -4a^2 \)

1. Rearrange the equation to set it equal to zero:
\[
6a^3 + 4a^2 - 42a = 0
\]

2. Factor out the greatest common factor (GCF), which is \( 2a \):
\[
2a(3a^2 + 2a - 21) = 0
\]

3. Solve for \( a \):
- From \( 2a = 0 \), we get \( a = 0 \).
- Solve the quadratic equation \( 3a^2 + 2a - 21 = 0 \) using the quadratic formula:
\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 3 \), \( b = 2 \), and \( c = -21 \):
\[
a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-21)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 252}}{6} = \frac{-2 \pm \sqrt{256}}{6} = \frac{-2 \pm 16}{6}
\]
This gives two solutions:
\[
a = \frac{-2 + 16}{6} = \frac{14}{6} = \frac{7}{3}, \quad a = \frac{-2 - 16}{6} = \frac{-18}{6} = -3
\]

4. The solutions are:
\[
a = 0, \quad a = \frac{7}{3}, \quad a = -3
\]

---

#### Equation 2: \( p^2 - 2p + 17 = 0 \)

1. Use the quadratic formula:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -2 \), and \( c = 17 \):
\[
p = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 17}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 68}}{2} = \frac{2 \pm \sqrt{-64}}{2}
\]
Since the discriminant is negative (\( \sqrt{-64} = 8i \)), the solutions are complex:
\[
p = \frac{2 \pm 8i}{2} = 1 \pm 4i
\]

2. The solutions are:
\[
p = 1 + 4i, \quad p = 1 - 4i
\]

---

#### Equation 3: \( r^2 - 8r - 15 = 0 \)

1. Factor the quadratic equation:
\[
r^2 - 8r - 15 = (r - 10)(r + 2) = 0
\]

2. Solve for \( r \):
\[
r - 10 = 0 \quad \text{or} \quad r + 2 = 0
\]
\[
r = 10, \quad r = -2
\]

3. The solutions are:
\[
r = 10, \quad r = -2
\]

---

#### Equation 4: \( (x^2 + 3)(x^2 - 2) = 0 \)

1. Set each factor equal to zero:
\[
x^2 + 3 = 0 \quad \text{or} \quad x^2 - 2 = 0
\]

2. Solve each equation:
- For \( x^2 + 3 = 0 \):
\[
x^2 = -3 \implies x = \pm \sqrt{-3} = \pm i\sqrt{3}
\]
- For \( x^2 - 2 = 0 \):
\[
x^2 = 2 \implies x = \pm \sqrt{2}
\]

3. The solutions are:
\[
x = \sqrt{2}, \quad x = -\sqrt{2}, \quad x = i\sqrt{3}, \quad x = -i\sqrt{3}
\]

---

#### Equation 5: \( 4x^2 - 100 = 0 \)

1. Factor out the GCF:
\[
4(x^2 - 25) = 0
\]

2. Recognize that \( x^2 - 25 \) is a difference of squares:
\[
x^2 - 25 = (x - 5)(x + 5)
\]

3. Solve for \( x \):
\[
x - 5 = 0 \quad \text{or} \quad x + 5 = 0
\]
\[
x = 5, \quad x = -5
\]

4. The solutions are:
\[
x = 5, \quad x = -5
\]

---

#### Equation 6: \( x^2 - 25x + 24 = 0 \)

1. Factor the quadratic equation:
\[
x^2 - 25x + 24 = (x - 24)(x - 1) = 0
\]

2. Solve for \( x \):
\[
x - 24 = 0 \quad \text{or} \quad x - 1 = 0
\]
\[
x = 24, \quad x = 1
\]

3. The solutions are:
\[
x = 24, \quad x = 1
\]

---

#### Equation 7: \( (x^2 - 6)(x - 1)(x + 1) = 0 \)

1. Set each factor equal to zero:
\[
x^2 - 6 = 0 \quad \text{or} \quad x - 1 = 0 \quad \text{or} \quad x + 1 = 0
\]

2. Solve each equation:
- For \( x^2 - 6 = 0 \):
\[
x^2 = 6 \implies x = \pm \sqrt{6}
\]
- For \( x - 1 = 0 \):
\[
x = 1
\]
- For \( x + 1 = 0 \):
\[
x = -1
\]

3. The solutions are:
\[
x = \sqrt{6}, \quad x = -\sqrt{6}, \quad x = 1, \quad x = -1
\]

---

#### Equation 8: \( x^4 - 14x^2 + 45 = 0 \)

1. Let \( u = x^2 \). Then the equation becomes:
\[
u^2 - 14u + 45 = 0
\]

2. Factor the quadratic equation:
\[
u^2 - 14u + 45 = (u - 9)(u - 5) = 0
\]

3. Solve for \( u \):
\[
u - 9 = 0 \quad \text{or} \quad u - 5 = 0
\]
\[
u = 9, \quad u = 5
\]

4. Substitute back \( u = x^2 \):
- For \( u = 9 \):
\[
x^2 = 9 \implies x = \pm 3
\]
- For \( u = 5 \):
\[
x^2 = 5 \implies x = \pm \sqrt{5}
\]

5. The solutions are:
\[
x = 3, \quad x = -3, \quad x = \sqrt{5}, \quad x = -\sqrt{5}
\]

---

#### Equation 9: \( 2p^4 - 27p^2 = -3p^3 \)

1. Rearrange the equation to set it equal to zero:
\[
2p^4 + 3p^3 - 27p^2 = 0
\]

2. Factor out the GCF, which is \( p^2 \):
\[
p^2(2p^2 + 3p - 27) = 0
\]

3. Solve for \( p \):
- From \( p^2 = 0 \), we get \( p = 0 \).
- Solve the quadratic equation \( 2p^2 + 3p - 27 = 0 \) using the quadratic formula:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = 3 \), and \( c = -27 \):
\[
p = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 + 216}}{4} = \frac{-3 \pm \sqrt{225}}{4} = \frac{-3 \pm 15}{4}
\]
This gives two solutions:
\[
p = \frac{-3 + 15}{4} = \frac{12}{4} = 3, \quad p = \frac{-3 - 15}{4} = \frac{-18}{4} = -\frac{9}{2}
\]

4. The solutions are:
\[
p = 0, \quad p = 3, \quad p = -\frac{9}{2}
\]

---

#### Equation 10: \( s^2 - s = 0 \)

1. Factor the equation:
\[
s(s - 1) = 0
\]

2. Solve for \( s \):
\[
s = 0 \quad \text{or} \quad s - 1 = 0
\]
\[
s = 0, \quad s = 1
\]

3. The solutions are:
\[
s = 0, \quad s = 1
\]

---

#### Equation 11: \( 3p^4 - 3p^2 = 10p^2 \)

1. Rearrange the equation to set it equal to zero:
\[
3p^4 - 3p^2 - 10p^2 = 0
\]
\[
3p^4 - 13p^2 = 0
\]

2. Factor out the GCF, which is \( p^2 \):
\[
p^2(3p^2 - 13) = 0
\]

3. Solve for \( p \):
- From \( p^2 = 0 \), we get \( p = 0 \).
- Solve the quadratic equation \( 3p^2 - 13 = 0 \):
\[
3p^2 = 13 \implies p^2 = \frac{13}{3} \implies p = \pm \sqrt{\frac{13}{3}}
\]

4. The solutions are:
\[
p = 0, \quad p = \sqrt{\frac{13}{3}}, \quad p = -\sqrt{\frac{13}{3}}
\]

---

#### Equation 12: \( 5x^2 - 2x = -1 \)

1. Rearrange the equation to set it equal to zero:
\[
5x^2 - 2x + 1 = 0
\]

2. Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 5 \), \( b = -2 \), and \( c = 1 \):
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot 1}}{2 \cdot 5} = \frac{2 \pm \sqrt{4 - 20}}{10} = \frac{2 \pm \sqrt{-16}}{10}
\]
Since the discriminant is negative (\( \sqrt{-16} = 4i \)), the solutions are complex:
\[
x = \frac{2 \pm 4i}{10} = \frac{1}{5} \pm \frac{2i}{5}
\]

3. The solutions are:
\[
x = \frac{1}{5} + \frac{2i}{5}, \quad x = \frac{1}{5} - \frac{2i}{5}
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
&1. \ a = 0, \ a = \frac{7}{3}, \ a = -3 \\
&2. \ p = 1 + 4i, \ p = 1 - 4i \\
&3. \ r = 10, \ r = -2 \\
&4. \ x = \sqrt{2}, \ x = -\sqrt{2}, \ x = i\sqrt{3}, \ x = -i\sqrt{3} \\
&5. \ x = 5, \ x = -5 \\
&6. \ x = 24, \ x = 1 \\
&7. \ x = \sqrt{6}, \ x = -\sqrt{6}, \ x = 1, \ x = -1 \\
&8. \ x = 3, \ x = -3, \ x = \sqrt{5}, \ x = -\sqrt{5} \\
&9. \ p = 0, \ p = 3, \ p = -\frac{9}{2} \\
&10. \ s = 0, \ s = 1 \\
&11. \ p = 0, \ p = \sqrt{\frac{13}{3}}, \ p = -\sqrt{\frac{13}{3}} \\
&12. \ x = \frac{1}{5} + \frac{2i}{5}, \ x = \frac{1}{5} - \frac{2i}{5}
\end{aligned}
}
\]