(x+3)(x-2)(x+5)=0 Standard Form

3 min read Jun 16, 2024

Solving a Cubic Equation in Standard Form

This article will walk you through the process of solving the cubic equation (x+3)(x-2)(x+5) = 0 and putting it in standard form.

Understanding the Problem

We have a cubic equation in factored form, meaning it is already expressed as the product of linear factors. To solve for x, we need to find the values of x that make the equation true, which means finding the values that make the product equal to zero.

The Zero Product Property

The Zero Product Property states that if the product of two or more factors is zero, then at least one of the factors must be zero. Applying this to our equation:

If (x+3) = 0, then x = -3
If (x-2) = 0, then x = 2
If (x+5) = 0, then x = -5

Therefore, the solutions to the equation are x = -3, x = 2, and x = -5.

Putting the Equation in Standard Form

Standard form for a cubic equation is ax³ + bx² + cx + d = 0. To achieve this, we need to expand the factored form:

Expand the first two factors: (x+3)(x-2) = x² + x - 6
Multiply the result by the remaining factor: (x² + x - 6)(x+5) = x³ + 6x² + x - 30

Therefore, the standard form of the equation is x³ + 6x² + x - 30 = 0.

Summary

We have solved the cubic equation (x+3)(x-2)(x+5) = 0 by:

Using the Zero Product Property to find the solutions: x = -3, x = 2, and x = -5.
Expanding the factored form to obtain the standard form: x³ + 6x² + x - 30 = 0.