## Algebra## Polynomial Equations |

## What are examples of polynomials with one root and no roots? |

The following is an example of a polynomial with only one root:

x^{2} + 6x + 9 = 0

(x^{2} + 6x + (6/2)^{2}) - (6/2)^{2} + 9 = 0

(x + 3)^{2} - 9 + 9 = 0

(x + 3) ^{2} = 0

x + 3 = 0

x = -3, or the polynomial has only one root *x* = -3

But not all polynomials have roots. The following is an example of a polynomial with no root:

2x^{2} - 6x + 8 = 0

(½)(2x^{2}- 6x + 8) = (½)0

x^{2} - 3x + 4 = 0

(x^{2} - 3x + (-3/2)^{2}) - (-3/2)^{2} + 4 = 0

(x - 3/2)^{2}- 9/4 + 4 = 0

(x - 3/2)^{2} + 7/4 = 0

(x - 3/2)^{2} = -7/4

Because a real number squared is greater than or equal to 0, that means (x - 3/2)^{2} will always be greater than or equal to 0. Therefore, the answer can’t be -7/4, a negative number, and there are no real roots for this polynomial.