To solve the quadratic equation $3x^2 - 2x - 72 = 0$ by factorization, follow these steps:

Step 1: Multiply the coefficient of $x^2$ ($3$) with the constant term ($-72$).

$3 \times -72 = -216$

Step 2: Find two numbers whose product is $-216$ and sum is the coefficient of $x$ ($-2$).

The numbers are $12$ and $-18$, because: $12 \times -18 = -216 \quad \text{and} \quad 12 + (-18) = -6$

Step 3: Rewrite the middle term ($-2x$) using these two numbers.

$3x^2 + 12x - 18x - 72 = 0$

Step 4: Group terms and factorize each group.

$(3x^2 + 12x) - (18x + 72) = 0$ $3x(x + 4) - 18(x + 4) = 0$

Step 5: Factor out the common binomial factor.

$(3x - 18)(x + 4) = 0$

Step 6: Solve for $x$.

Set each factor equal to $0$: $3x - 18 = 0 \quad \Rightarrow \quad x = \frac{18}{3} = 6$ $x + 4 = 0 \quad \Rightarrow \quad x = -4$

Final Solution:

The solutions are: $x = 6 \quad \text{or} \quad x = -4$

Would you like a detailed explanation of any step?

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Related Questions:

1. How do you verify the solutions of a quadratic equation?
2. What are other methods to solve quadratic equations, such as completing the square or using the quadratic formula?
3. How can you determine if a quadratic equation can be solved by factorization?
4. Can this factorization method be applied to cubic equations? If not, why?
5. What are the geometric interpretations of the solutions of a quadratic equation?

Tip:

Always check if the quadratic equation can be factored easily by inspecting its coefficients and possible integer roots before applying complex methods!