- Identify the greatest common factor (GCF) of the polynomial terms as \(7x^2\).
- Factor out \(7x^2\) to get \(7x^2(x^2 + 2x - 24)\).
- Factor the remaining quadratic trinomial \(x^2 + 2x - 24\) into \((x - 4)(x + 6)\) by finding two numbers with a sum of 2 and a product of -24.
- Combine the results to get the final factored form \(\boxed{7x^2(x - 4)(x + 6)}\).
Explanation
- Identifying the Greatest Common Factor
To find the completely factored form of the polynomial \(7x^4 + 14x^3 - 168x^2\), we start by identifying the greatest common factor (GCF) of all the terms. First, look at the coefficients: 7, 14, and -168. Since \(7 \times 1 = 7\), \(7 \times 2 = 14\), and \(7 \times (-24) = -168\), the common numerical factor is 7. Next, examine the variables: \(x^4\), \(x^3\), and \(x^2\). The highest power of \(x\) that divides all these terms is the lowest exponent present, which is \(x^2\). Combining these, the GCF of the entire expression is \(7x^2\).
- Factoring the Quadratic Trinomial
Now, we factor out the GCF \(7x^2\) from the original polynomial:
\[7x^4 + 14x^3 - 168x^2 = 7x^2(x^2 + 2x - 24).\]
We are left with a quadratic trinomial, \(x^2 + 2x - 24\), which we need to factor further. To factor this, we look for two numbers that multiply to the constant term (-24) and add up to the coefficient of the middle linear term (2). Listing the factors of -24:
- \(-1 \times 24 = -24, -1+24 = 23\)
- \(1 \times -24 = -24, 1-24 = -23\)
- \(-2 \times 12 = -24, -2+12 = 10\)
- \(2 \times -12 = -24, 2-12 = -10\)
- \(-3 \times 8 = -24, -3+8 = 5\)
- \(3 \times -8 = -24, 3-8 = -5\)
- \(-4 \times 6 = -24, -4+6 = 2\)
The pair \(-4\) and \(6\) satisfies both conditions. Therefore, the trinomial factors into \((x - 4)(x + 6)\).
- Conclusion and Final Result
By putting it all together, we have:
\[7x^4 + 14x^3 - 168x^2 = 7x^2(x - 4)(x + 6).\]
Comparing this result with the given choices, we see that it matches the first option. Thus, the completely factored form of the polynomial is \(7x^2(x - 4)(x + 6)\).
Examples
Factoring polynomials is a fundamental skill in algebra with many real-world applications. For instance, when designing a rectangular garden bed, the area might be represented by a polynomial expression. By factoring the expression, you can determine the algebraic representations of the length and width of the garden. This helps in calculating materials needed for fencing, such as the perimeter \(P = 2 \times (\text{length} + \text{width})\), directly from the factored dimensions. In physics and engineering, factoring is often used to solve for equilibrium points in a system by setting the expression equal to zero and finding its roots, which are easily identified once the expression is in factored form.