(x-2)(x-5).jpeg "(x+3)(x-2)(x-5)")
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.