Absolutely! Let's solve each of these equations step by step using factorization.
### (3) [tex]\( x^2 - 5x = 0 \)[/tex]
To solve this equation using factorization, follow these steps:
1. Rewrite the equation:
[tex]\[
x^2 - 5x = 0
\][/tex]
2. Factor out the common term [tex]\( x \)[/tex]:
[tex]\[
x(x - 5) = 0
\][/tex]
3. Set each factor equal to zero:
[tex]\[
x = 0 \quad \text{or} \quad x - 5 = 0
\][/tex]
4. Solve each equation:
[tex]\[
x = 0 \quad \text{or} \quad x = 5
\][/tex]
So, the solutions are:
[tex]\[
x = 0 \quad \text{and} \quad x = 5
\][/tex]
### 2. [tex]\( a^2 x^2 - b^2 = 0 \)[/tex]
To solve this equation using factorization, follow these steps:
1. Rewrite the equation as a difference of squares:
[tex]\[
a^2 x^2 - b^2 = 0
\][/tex]
2. Use the difference of squares formula [tex]\( A^2 - B^2 = (A - B)(A + B) \)[/tex], where [tex]\( A = ax \)[/tex] and [tex]\( B = b \)[/tex]:
[tex]\[
(ax - b)(ax + b) = 0
\][/tex]
3. Set each factor equal to zero:
[tex]\[
ax - b = 0 \quad \text{or} \quad ax + b = 0
\][/tex]
4. Solve each equation:
[tex]\[
ax = b \quad \Rightarrow \quad x = \frac{b}{a}
\][/tex]
[tex]\[
ax = -b \quad \Rightarrow \quad x = \frac{-b}{a}
\][/tex]
So, the solutions are:
[tex]\[
x = \frac{b}{a} \quad \text{and} \quad x = \frac{-b}{a}
\][/tex]
### 3. [tex]\( bx^2 + abx = 0 \)[/tex]
To solve this equation using factorization, follow these steps:
1. Rewrite the equation:
[tex]\[
bx^2 + abx = 0
\][/tex]
2. Factor out the common term [tex]\( bx \)[/tex]:
[tex]\[
bx(x + a) = 0
\][/tex]
3. Set each factor equal to zero:
[tex]\[
bx = 0 \quad \text{or} \quad x + a = 0
\][/tex]
4. Solve each equation:
[tex]\[
bx = 0 \quad \Rightarrow \quad x = 0 \quad (\text{since } b \neq 0)
\][/tex]
[tex]\[
x + a = 0 \quad \Rightarrow \quad x = -a
\][/tex]
So, the solutions are:
[tex]\[
x = 0 \quad \text{and} \quad x = -a
\][/tex]
By following these steps, we have successfully solved the given equations using the factorization method.
