Three functions: $f(x)$, $g(x)$ and $h(x)$ are defined as follows: $$f(x) = 2x^2 + x - 6$$ $$g(x) = x^3 - 6x + 9$$ $$h(x) = x^2 - 2x.$$ Solve $f(x) = 0$ - Junior Cycle Mathematics - Question a - 2013

For the equation $g(x) = 0$, we have:

$x^3 - 6x + 9 = 0.$

To find the roots, we can apply the Factor Theorem. Testing $x = 3$:

$g(3) = 3^3 - 6(3) + 9 = 27 - 18 + 9 = 18.$

Next, we perform polynomial division on $x^3 - 6x + 9$ by $(x - 3)$ to factor it:

After successfully factoring, we have:

$(x - 3)(x^2 + 3) = 0.$

The quadratic $x^2 + 3 = 0$ gives us no real solutions but the factor $x - 3$ provides:

$x = 3.$

Thus, the solution is: $x = 3$.

For the equation $h(x) = 0$, we write:

$x^2 - 2x = 0.$

Factoring yields:

$x(x - 2) = 0.$

Setting each factor to zero gives the solution:

1. $x = 0$
2. $x = 2$

Thus, the solutions are: $x = 0$ and $x = 2$.