Question

Consider this polynomial equation.

Use the equation to complete this statement.
The equation has solutions. Its real solutions are .

Answer 1

x = 3, -1, 2, -2

Explanation:
`x^(2)`

6(
`x`-3)(
`x^(2)`+4)(
`x`+1) = 0

6(
`x`-3)(
`x`+2)(
`x`-2)(
`x`+1) = 0

Using the fact that (
`x^(2)`+4) = (
`x`-2)(
`x`+2) - known as the difference of two squares rule

Thus solve by inserting values of
`x` that will result in the equation equating to 0.

ie: if
`x` = 3,

6(3-3)(3+2)(3-2)(3+1) = 0

0=0

RHS = LHS therefore true

Answer 2

The equation has 5 solutions and its real soluations are x=3, +sqrt2, -sqrt2, and -1

Explanation: