Let's solve each quadratic equation by factoring.
### Equation 1: [tex]\(x^2 + 10 = 0\)[/tex]
First, we need to factor the equation. Notice that [tex]\(x^2 + 10\)[/tex] does not factor nicely with real numbers since the term [tex]\(10\)[/tex] cannot be split into real numbers that sum up to a factorable quadratic. However, we can consider its factored form as it stands:
Factored Form:
[tex]\[ x^2 + 10 \][/tex]
Now, to find the solutions, set the equation to zero:
[tex]\[ x^2 + 10 = 0 \][/tex]
Subtract 10 from both sides:
[tex]\[ x^2 = -10 \][/tex]
Take the square root of both sides. Remember that the square root of a negative number involves imaginary numbers (i):
[tex]\[ x = \pm \sqrt{-10} \][/tex]
[tex]\[ x = \pm \sqrt{10}i \][/tex]
Thus, the solutions are:
[tex]\[ x = -\sqrt{10}i \quad \text{and} \quad x = \sqrt{10}i \][/tex]
### Equation 2: [tex]\(4x^2 + 25 = 0\)[/tex]
Next, let's factor the second equation. Similar to the first equation, [tex]\(4x^2 + 25\)[/tex] does not factor nicely with real numbers but we will write its factored form:
Factored Form:
[tex]\[ 4x^2 + 25 \][/tex]
Now, set the equation to zero:
[tex]\[ 4x^2 + 25 = 0 \][/tex]
Subtract 25 from both sides:
[tex]\[ 4x^2 = -25 \][/tex]
Divide both sides by 4:
[tex]\[ x^2 = -\frac{25}{4} \][/tex]
Take the square root of both sides:
[tex]\[ x = \pm \sqrt{ -\frac{25}{4} } \][/tex]
Simplify the square root (taking into account the imaginary unit [tex]\(i\)[/tex]):
[tex]\[ x = \pm \frac{5i}{2} \][/tex]
Therefore, the solutions are:
[tex]\[ x = -\frac{5i}{2} \quad \text{and} \quad x = \frac{5i}{2} \][/tex]
### Summary
Now completing the table with the factors and solutions, we get:
\begin{tabular}{|l|l|l|}
\hline Equation & Factors & Solutions \\
\hline [tex]$x^2 + 10 = 0$[/tex] & [tex]\(x^2 + 10\)[/tex] & [tex]\([- \sqrt{10}i, \sqrt{10}i]\)[/tex] \\
\hline [tex]$4x^2 + 25 = 0$[/tex] & [tex]\(4x^2 + 25\)[/tex] & [tex]\([- \frac{5i}{2}, \frac{5i}{2}]\)[/tex] \\
\hline
\end{tabular}
