- The polynomial \(P(x) = 2x^3 - 3x^2 - 3x + 2\) is divided by \((x - 2)\) using polynomial division.
- The resulting quotient is \(Q(x) = 2x^2 + x - 1\).
- Factoring the quadratic \(Q(x)\) yields \((2x - 1)(x + 1)\).
- Comparing with the given choices, the correct answer is \(\boxed{(2x - 1)(x + 1)}\).
Explanation
- Analyzing the Polynomial and Dividing by the Known Factor
The given polynomial is \(P(x) = 2x^3 - 3x^2 - 3x + 2\). We are given that \((x - 2)\) is a factor, which means the polynomial can be expressed as the product of this factor and a quadratic expression \(Q(x)\). To find \(Q(x)\), we perform polynomial long division of \(P(x)\) by \((x - 2)\).
- Performing Polynomial Division
Dividing \(2x^3 - 3x^2 - 3x + 2\) by \((x - 2)\):
- First, we divide the leading term \(2x^3\) by \(x\) to get \(2x^2\).
- Multiply \(2x^2\) by \((x - 2)\) to get \(2x^3 - 4x^2\).
- Subtract this from the original polynomial: \((2x^3 - 3x^2 - 3x + 2) - (2x^3 - 4x^2) = x^2 - 3x + 2\).
- Next, divide the new leading term \(x^2\) by \(x\) to get \(x\).
- Multiply \(x\) by \((x - 2)\) to get \(x^2 - 2x\).
- Subtract this: \((x^2 - 3x + 2) - (x^2 - 2x) = -x + 2\).
- Finally, divide \(-x\) by \(x\) to get \(-1\).
- Multiply \(-1\) by \((x - 2)\) to get \(-x + 2\), leaving a remainder of 0.
Thus, the quotient is \(Q(x) = 2x^2 + x - 1\).
- Factoring the Resulting Quadratic Expression
We now need to factor the quadratic expression \(Q(x) = 2x^2 + x - 1\). We look for two factors of the form \((ax + b)(cx + d)\) that result in \(2x^2 + x - 1\). We can see that \((2x - 1)(x + 1)\) expands to \(2x^2 + 2x - x - 1 = 2x^2 + x - 1\). Therefore, the factored form of the quadratic quotient is \((2x - 1)(x + 1)\).
- Final Conclusion and Selecting the Correct Answer
Comparing our finding \((2x - 1)(x + 1)\) with the given options, we see that it perfectly matches the last option. Therefore, the expression representing the other factors of the polynomial is \((2x - 1)(x + 1)\).
Examples
Polynomial factorization is a crucial skill in algebra, used in many real-world applications such as physics simulations and engineering design. For instance, when modeling the motion of a projectile, the path of the object can often be represented by a quadratic or cubic polynomial. By factoring these polynomials, engineers can easily determine critical points like the ground impact time, where the position equals zero (the roots of the polynomial). This helps in predicting trajectories accurately and ensuring safety in systems like automated landing gear.