(x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0

Jasonbradley

4 min read Jun 17, 2024

Solving the Equation: (x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0

This article will guide you through the steps of solving the given equation.

Expanding the Equation

First, let's expand the equation to simplify it:

Expand the first term: (x-2)(x^2+2x+7) = x^3 + 2x^2 + 7x - 2x^2 - 4x - 14 = x^3 + 3x - 14
Expand the second term: 2(x^2-4) = 2x^2 - 8
Expand the third term: -5(x-2) = -5x + 10

Now the equation becomes: x^3 + 3x - 14 + 2x^2 - 8 - 5x + 10 = 0

Combining Like Terms

Next, combine the like terms: x^3 + 2x^2 - 2x - 12 = 0

Factoring the Equation

Now, we need to factor the equation to solve for x. Unfortunately, this cubic equation doesn't factor easily. We can use the Rational Root Theorem to test for possible rational roots:

Rational Root Theorem: If a polynomial equation has integer coefficients, then any rational root must be of the form p/q, where 'p' is a factor of the constant term and 'q' is a factor of the leading coefficient.

In our equation:

Constant term: -12
Leading coefficient: 1

Therefore, the possible rational roots are the factors of -12 ( ±1, ±2, ±3, ±4, ±6, ±12) divided by the factors of 1 ( ±1).

We can test each of these possible roots by substituting them into the equation. We find that x = 2 is a root of the equation.

Using Synthetic Division

Now that we know x=2 is a root, we can use synthetic division to factor the equation:

2 | 1  2  -2  -12
    | 2  8  12
    ------------------
      1  4   6    0

This gives us a new equation: (x-2)(x^2 + 4x + 6) = 0

The quadratic factor (x^2 + 4x + 6) doesn't factor easily. We can use the quadratic formula to find the remaining roots:

Quadratic Formula

The quadratic formula is: x = (-b ± √(b^2 - 4ac)) / 2a

Where a = 1, b = 4, and c = 6 in our quadratic factor.

Solving for x, we get:

x = (-4 ± √(4^2 - 4 * 1 * 6)) / (2 * 1)

x = (-4 ± √(-8)) / 2

x = (-4 ± 2√2i) / 2

x = -2 ± √2i

Solutions

Therefore, the solutions to the equation (x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0 are:

x = 2
x = -2 + √2i
x = -2 - √2i

These are the three roots of the cubic equation, one real root and two complex roots.