- Divide the polynomial \(x^4 + 2x^3 - 7x^2 - 8x + 12\) by the known factor \((x-2)\) to obtain the cubic polynomial \(q(x) = x^3 + 4x^2 + x - 6\).
- Find the roots of the cubic polynomial by testing potential values, discovering that \(x=1\) is a root, identifying \((x-1)\) as a factor.
- Factor the remaining quadratic \(x^2 + 5x + 6\) into \((x+2)(x+3)\).
- Conclude that the other three factors are \(\boxed{x-1, x+2, x+3}\).
Explanation
- Problem Analysis
We are given the polynomial \(p(x) = x^4 + 2x^3 - 7x^2 - 8x + 12\) and we are told that \((x - 2)\) is a factor. Our goal is to determine the other three linear factors of this polynomial. Since the degree of the polynomial is 4, we expect to find four linear factors in total, one of which is already known. Thus, we can write \(p(x) = (x - 2) \cdot q(x)\), where \(q(x)\) is a cubic polynomial.
- Polynomial Division
To find the cubic polynomial \(q(x)\), we divide \(p(x) = x^4 + 2x^3 - 7x^2 - 8x + 12\) by \((x - 2)\). Performing polynomial long division (or synthetic division):
Dividing \(x^4\) by \(x\) gives \(x^3\).
Multiplying \(x^3(x-2) = x^4 - 2x^3\).
Subtracting \((x^4 + 2x^3 - 7x^2 - 8x + 12) - (x^4 - 2x^3) = 4x^3 - 7x^2 - 8x + 12\).
Dividing \(4x^3\) by \(x\) gives \(4x^2\).
Multiplying \(4x^2(x-2) = 4x^3 - 8x^2\).
Subtracting \((4x^3 - 7x^2 - 8x + 12) - (4x^3 - 8x^2) = x^2 - 8x + 12\).
Dividing \(x^2\) by \(x\) gives \(x\).
Multiplying \(x(x-2) = x^2 - 2x\).
Subtracting \((x^2 - 8x + 12) - (x^2 - 2x) = -6x + 12\).
Dividing \(-6x\) by \(x\) gives \(-6\).
Multiplying \(-6(x-2) = -6x + 12\).
The remainder is 0, confirming \((x-2)\) is a factor. Thus, \(q(x) = x^3 + 4x^2 + x - 6\).
- Factoring the Cubic Polynomial
Now we need to factor the cubic polynomial \(q(x) = x^3 + 4x^2 + x - 6\). By testing small integer roots (factors of the constant term \(-6\)), we test \(x=1\):
\(q(1) = 1^3 + 4(1)^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0\).
Since \(q(1) = 0\), \((x - 1)\) is a factor. Dividing \(q(x)\) by \((x - 1)\) gives:
\(x^3 + 4x^2 + x - 6 = (x - 1)(x^2 + 5x + 6)\).
Now we factor the quadratic \(x^2 + 5x + 6\). We look for two numbers that multiply to 6 and add to 5, which are 2 and 3.
Thus, \(x^2 + 5x + 6 = (x + 2)(x + 3)\).
The complete factorization of \(p(x)\) is \((x - 2)(x - 1)(x + 2)(x + 3)\). The other three factors are \((x-1)\), \((x+2)\), and \((x+3)\).
Examples
Polynomial factorization is a fundamental tool in engineering and physics, where it helps in simplifying complex differential equations and understanding system stability. For example, when analyzing the vibration modes of a bridge or the signal response in an electrical circuit, engineers often end up with high-degree polynomials that must be factored to find the system's "poles" or natural frequencies. Factoring \(p(x) = x^4 + 2x^3 - 7x^2 - 8x + 12\) into \((x - 2)(x - 1)(x + 2)(x + 3)\) reveals that the system's critical points are at \(x = 2, 1, -2, -3\), allowing engineers to predict how the system will behave under various conditions.