Use factoring to solve the polynomial equation:
$2 x^2=2-3 x$
The given equation is \(2x^2 = 2 - 3x\). To solve this quadratic equation, we first bring all terms to one side of the equation to put it in the standard form \(ax^2 + bx + c = 0\). Adding \(3x\) and subtracting \(2\) from both sides gives:
\[2x^2 + 3x - 2 = 0\]
This is a quadratic equation where \(a=2, b=3, c=-2\).
To factor the quadratic expression \(2x^2 + 3x - 2\), we need to find two numbers that multiply to the product of the leading coefficient and the constant term, which is \(a \times c = 2 \times (-2) = -4\), and whose sum is equal to the coefficient of the linear term, \(b = 3\).
Looking at the factors of \(-4\), we find that \(-1\) and \(4\) satisfy both conditions:
\[(-1) \times 4 = -4\]
\[(-1) + 4 = 3\]
Now, we rewrite the middle term \(3x\) using these two numbers:
\[2x^2 + 4x - 1x - 2 = 0\]
We group the terms to factor by grouping:
\[(2x^2 + 4x) + (-1x - 2) = 0\]
Factoring out the greatest common factor from each group, we get:
\[2x(x + 2) - 1(x + 2) = 0\]
Now, we can factor out the common binomial \((x + 2)\):
\[(2x - 1)(x + 2) = 0\]
For the product of two factors to be zero, at least one of the factors must be zero. Setting each factor equal to zero gives us:
1) \(2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}\)
2) \(x + 2 = 0 \implies x = -2\)
Thus, the solutions to the equation are \(x = \frac{1}{2}\) and \(x = -2\). These values satisfy the original equation \(2x^2 = 2 - 3x\). The final answer is \(\boxed{x = \frac{1}{2}, x = -2}\).
In physics, quadratic equations often arise when modeling the motion of objects under constant acceleration. For example, if you throw a ball into the air, its height as a function of time can be represented by a quadratic expression. Solving this equation helps determine exactly when the ball reaches a certain height or hits the ground. By using techniques like factoring, you can quickly find these specific time values in real-world scenarios.