(x+1)(x+3)(x+5)(x+7)+15=0

4 min read Jun 16, 2024

Solving the Equation (x+1)(x+3)(x+5)(x+7) + 15 = 0

This equation looks complicated, but we can solve it by using some clever algebraic manipulations and recognizing patterns.

Step 1: Expanding the Equation

First, let's expand the product of the first four terms:

(x+1)(x+3)(x+5)(x+7) = (x² + 4x + 3)(x² + 12x + 35)

Now, we need to expand this further. You can use the distributive property or a more visual method like the FOIL method to do this:

(x² + 4x + 3)(x² + 12x + 35) = x⁴ + 16x³ + 83x² + 164x + 105

Therefore, the original equation becomes:

x⁴ + 16x³ + 83x² + 164x + 105 + 15 = 0

Simplifying, we get:

x⁴ + 16x³ + 83x² + 164x + 120 = 0

Step 2: Recognizing a Pattern

Now, let's try to factor the equation. Notice that the constant term (120) can be factored as 1 x 2 x 3 x 4 x 5. This looks suspiciously similar to the constant terms in the original factors (1, 3, 5, 7).

Let's try grouping the terms:

(x⁴ + 16x³ + 83x² + 164x) + 120 = 0

Now, we can factor out a common factor of x from the first group:

x(x³ + 16x² + 83x + 164) + 120 = 0

We can further factor the expression inside the parentheses. Notice that the coefficients follow a pattern: 1, 16, 83, 164. If we divide each coefficient by the previous one, we get a pattern of 16, 5.1875, 1.976, which suggests a possible quadratic factor.

Using the Rational Root Theorem, we can try potential rational roots of the cubic expression. After some trials, we find that x = -4 is a root. This means (x + 4) is a factor.

Dividing the cubic expression (x³ + 16x² + 83x + 164) by (x + 4) using polynomial long division, we get:

x³ + 16x² + 83x + 164 = (x + 4)(x² + 12x + 41)

Therefore, our equation becomes:

x(x + 4)(x² + 12x + 41) + 120 = 0

Step 3: Solving the Quadratic Factor

The quadratic factor (x² + 12x + 41) doesn't factor easily. We can solve for its roots using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

where a = 1, b = 12, and c = 41.

Substituting the values, we get:

x = (-12 ± √(12² - 4 * 1 * 41)) / 2 * 1 x = (-12 ± √(-12)) / 2 x = (-12 ± 2√3i) / 2 x = -6 ± √3i

Step 4: Final Solutions

Therefore, the solutions to the equation (x+1)(x+3)(x+5)(x+7) + 15 = 0 are:

x = -4
x = -6 + √3i
x = -6 - √3i

We have two real solutions and two complex solutions.