First we want to find the factors of the constant term - 55 then choose the pair of factors (let's call them p and q) that add up to 6. After finding p and q, we factor the quadratic by writing it in the form \left(x + p\right)\left(x + q\right)=0.

The prime factors of -55 are \pm1,\, \pm 5,\, \pm11 and \pm 55, and we want a factor pair whose sum is 6.

The factor pair whose sum is 6 is -5 and 11, so we have p=-5, q=11.

Factoring gives us x^2+6x-55=\left(x-5\right)\left(x+11\right) which leads to the equation \left(x-5\right)\left(x+11\right)=0

We can then solve the equation by setting each factor equal to zero, giving us x-5=0 and x+11=0, which gives the solutions x=5 and x=-11.

It is important to pay attention to the sign of the terms when determining the solution from a quadratic in factored form. In the example above, the term of \left(x-5\right) leads to a solution of x=5 and \left(x+11\right) to a solution of x=-11.

It is also important to note that in the two forms \left(x+p\right)\left(x+q\right) and \left(x-x_1\right)\left(x-x_2\right), we can see that x_1=-p, and x_2=-q.

To write the equation for this quadratic in factored form we need to first identify the solutions of the equation. We can then substitute these values for x_1 and x_2. Next, we will need to find a by substituting one other point and solving the resulting equation.

The roots of the equation are at \left(-1,\,0\right) and \left(3,\,0\right) so we know the equation has solutions of -1 and 3.

Since the solutions are -1 and 3, we can write the factored form as:

y=a(x+1)(x-3)

for some value of a. We can find a by substituting in the coordinates of the additional point, \left(4,\,10\right), into the function.

To find a:

The equation of the quadratic function in factored form:

y=2(x+1)(x-3)

We can apply the Factor theorem to identify the binomial factors of f\left(x\right) and then identify if there is a constant factor a.

The factor theorem states that if f\left(a\right)=0 then \left(x-a\right) must be a factor of f\left(x\right).

Since f\left(2\right)=0 then \left(x-2\right) must be a factor of f\left(x\right). And since f\left(-5\right)=0 then \left(x-\left(-5\right)\right) or \left(x+5\right) must be a factor of f\left(x\right).

This tells us that f\left(x\right)=a\left(x-2\right)\left(x+5\right) where a is a constant factor of f\left(x\right). Now we must determine if f\left(x\right) has a constant factor by using the fact that f\left(4\right)=2.

Since we know that a=\dfrac{1}{9} we can substitute it for a giving us the equation f\left(x\right)=\dfrac{1}{9}\left(x-2\right)\left(x+5\right).