(x+1)(2x+3)(2x+5)(x+3)=945 Quadratic

5 min read Jun 16, 2024

Solving the Quadratic Equation: (x+1)(2x+3)(2x+5)(x+3) = 945

This problem involves expanding and simplifying a product of four linear factors, resulting in a fourth-degree polynomial equation. However, we can use strategic manipulation and factoring to solve this equation efficiently.

Step 1: Simplify the Equation

First, we need to expand the product on the left-hand side of the equation:

(x+1)(2x+3)(2x+5)(x+3) = 945

Let's start by multiplying the first two factors and the last two factors:

(2x² + 5x + 3)(2x² + 11x + 15) = 945

Next, we need to multiply these two quadratic expressions. This can be done by using the distributive property or by setting up a grid method:

4x⁴ + 22x³ + 30x² + 10x³ + 55x² + 75x + 6x² + 33x + 45 = 945

Combining like terms, we get:

4x⁴ + 32x³ + 91x² + 108x - 900 = 0

Step 2: Finding Rational Roots

Now we have a fourth-degree polynomial equation. We can try to find rational roots using the Rational Root Theorem. This theorem states that any rational root of a polynomial with integer coefficients must be of the form p/q, where p is a factor of the constant term (in our case, -900) and q is a factor of the leading coefficient (in our case, 4).

The factors of -900 are ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±25, ±30, ±36, ±45, ±50, ±60, ±75, ±90, ±100, ±150, ±180, ±225, ±300, ±450, ±900.

The factors of 4 are ±1, ±2, ±4.

We can now test all possible combinations of p/q. After some trial and error, we find that x = 3 is a root of the equation.

Step 3: Using the Factor Theorem

Since x = 3 is a root, we know that (x - 3) is a factor of the polynomial. We can use polynomial division or synthetic division to divide the polynomial by (x - 3). This will give us a cubic polynomial:

4x⁴ + 32x³ + 91x² + 108x - 900 = (x - 3)(4x³ + 44x² + 199x + 300)

Step 4: Factoring the Cubic Polynomial

Now we have to factor the cubic polynomial:

4x³ + 44x² + 199x + 300

We can try to factor this by grouping or by using the Rational Root Theorem again. However, in this case, it is easier to observe that x = -3 is also a root of the cubic polynomial. Therefore, (x + 3) is a factor:

4x³ + 44x² + 199x + 300 = (x + 3)(4x² + 32x + 100)

Step 5: Solving the Quadratic Equation

Now we are left with a quadratic equation:

4x² + 32x + 100 = 0

We can solve this quadratic equation by using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

Where a = 4, b = 32, and c = 100.

Substituting these values into the quadratic formula, we get:

x = (-32 ± √(32² - 4 * 4 * 100)) / (2 * 4)

x = (-32 ± √(-336)) / 8

x = (-32 ± 4√(21)) / 8

x = -4 ± (√21)/2

Solutions

Therefore, the solutions to the equation (x+1)(2x+3)(2x+5)(x+3)=945 are:

x = 3
x = -3
x = -4 + (√21)/2
x = -4 - (√21)/2

These are the four solutions to the given quadratic equation.