(x-1)(x+2)(x+3)(x+6)-28

7 min read Jun 17, 2024

Factoring and Solving the Expression (x-1)(x+2)(x+3)(x+6) - 28

This article will explore the process of factoring and solving the expression (x-1)(x+2)(x+3)(x+6) - 28. We will break down the steps involved and illustrate the key concepts used.

Step 1: Simplifying the Expression

The first step is to simplify the expression by expanding the product of the four linear factors. This can be done in a systematic way:

Group the factors: Group the first two factors and the last two factors together: [(x-1)(x+2)][(x+3)(x+6)] - 28.
Expand the groups: Use the FOIL method (First, Outer, Inner, Last) to expand each group: [(x² + x - 2)][(x² + 9x + 18)] - 28.
Expand further: Apply the FOIL method again to expand the product of the two quadratic expressions: x⁴ + 10x³ + 23x² + 12x - 36 - 28.

Now, we have a simplified expression: x⁴ + 10x³ + 23x² + 12x - 64.

Step 2: Factoring the Expression

The next step is to factor the simplified expression. This can be a challenging task, but we can use a few strategies:

Look for common factors: In this case, there are no common factors for all the terms.
Try grouping: We can try grouping the terms in pairs to see if any common factors emerge. However, this method doesn't yield useful factors in this specific expression.
Rational Root Theorem: The Rational Root Theorem can help us find potential rational roots of the polynomial. This theorem states that any rational root of a polynomial must be of the form p/q, where p is a factor of the constant term (in our case, -64) and q is a factor of the leading coefficient (in our case, 1).
- Factors of -64: ±1, ±2, ±4, ±8, ±16, ±32, ±64
- Factors of 1: ±1
Therefore, the potential rational roots are: ±1, ±2, ±4, ±8, ±16, ±32, ±64.
Synthetic Division: We can use synthetic division to test the potential rational roots found in the previous step. If a potential root leads to a remainder of zero, then it is a root of the polynomial. By performing synthetic division with the potential roots, we find that x = 1 is a root of the polynomial.
Factoring with the Root: Knowing that x = 1 is a root, we can factor the polynomial using the Factor Theorem: (x - 1) is a factor of the polynomial. Performing polynomial long division or synthetic division again, we can obtain the remaining factor: (x - 1)(x³ + 11x² + 34x + 64).
Factoring the Cubic: Factoring the cubic expression (x³ + 11x² + 34x + 64) can be challenging. We can try using the Rational Root Theorem and synthetic division again. After testing, we find that x = -4 is a root of this cubic expression.
Further Factoring: Factoring (x³ + 11x² + 34x + 64) with the root x = -4, we get: (x + 4)(x² + 7x + 16).

Finally, we have factored the original expression: (x - 1)(x + 4)(x² + 7x + 16).

Step 3: Solving the Equation

The expression we obtained is equal to zero. We can find the solutions by setting each factor equal to zero and solving for x:

(x - 1) = 0 => x = 1
(x + 4) = 0 => x = -4
(x² + 7x + 16) = 0 => This quadratic factor does not factor easily. We can use the quadratic formula to find the solutions:
- x = (-b ± √(b² - 4ac)) / 2a
- x = (-7 ± √(7² - 4 * 1 * 16)) / 2 * 1
- x = (-7 ± √(-15)) / 2
- x = (-7 ± √15i) / 2 (where i is the imaginary unit, √-1)

Therefore, the solutions to the equation (x-1)(x+2)(x+3)(x+6)-28 = 0 are x = 1, x = -4, x = (-7 + √15i) / 2, and x = (-7 - √15i) / 2.

Conclusion

By using simplification, factoring, and solving techniques, we successfully factored and solved the expression (x-1)(x+2)(x+3)(x+6)-28. This process highlights the importance of understanding fundamental algebraic concepts and applying them systematically to address complex expressions.