## Factoring and Solving the Polynomial (x+3)(x-2)(x-5)

This article explores the factored form of the polynomial (x+3)(x-2)(x-5) and its implications for finding its roots (or zeros).

### Understanding the Factored Form

The expression (x+3)(x-2)(x-5) is already presented in its factored form. This means the polynomial is expressed as the product of several linear factors. Each factor represents a linear expression, which is an expression of the form (x - a), where 'a' is a constant.

### Finding the Roots

The roots of a polynomial are the values of 'x' that make the polynomial equal to zero. To find the roots of the factored polynomial (x+3)(x-2)(x-5), we can use the following principle:

**The product of factors is zero if and only if at least one of the factors is zero.**

Therefore, to find the roots of the polynomial, we set each factor equal to zero and solve for 'x':

**x + 3 = 0**--> x = -3**x - 2 = 0**--> x = 2**x - 5 = 0**--> x = 5

Thus, the roots of the polynomial (x+3)(x-2)(x-5) are **-3, 2, and 5**.

### Expanding the Polynomial

While the factored form is useful for finding the roots, we can also expand the polynomial to get its standard 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 - 4x² - 11x + 30

Therefore, the expanded form of the polynomial is **x³ - 4x² - 11x + 30**.

### Conclusion

The polynomial (x+3)(x-2)(x-5) is presented in its factored form, making it easy to determine its roots, which are -3, 2, and 5. The factored form also allows us to expand the polynomial to its standard form, x³ - 4x² - 11x + 30.