To solve the quadratic equation \( x^2 + 2x - 80 = 0 \) by factoring, we first need to find two numbers that multiply to \(-80\) (the constant term) and add up to \(2\) (the coefficient of the \(x\) term).
The two numbers that satisfy these conditions are \(10\) and \(-8\) since:
- \(10 \times -8 = -80\)
- \(10 + (-8) = 2\)
Now we can rewrite the quadratic equation using these two numbers:
\[ x^2 + 10x - 8x - 80 = 0 \]
Next, we can group the terms:
\[ (x^2 + 10x) + (-8x - 80) = 0 \]
Factoring out the common terms in each group gives:
\[ x(x + 10) - 8(x + 10) = 0 \]
Now, factor out \((x + 10)\):
\[ (x + 10)(x - 8) = 0 \]
To find the values of \(x\), we set each factor to zero:
- \(x + 10 = 0 \Rightarrow x = -10\)
- \(x - 8 = 0 \Rightarrow x = 8\)
Thus, the solutions to the equation \(x^2 + 2x - 80 = 0\) are:
\[ \boxed{-10} \quad \text{and} \quad \boxed{8} \]
1 answer