One can remember that if the multiplicity of a zero is odd then it passes through the x-axis and if it's even then it 'bounces' off the x-axis. You can think about this analytically as well. What happens when we plug a number into our function just slightly above or below a zero with an even multiplicity? You find that the sign is always positive. Whereas a zero with an odd multiplicity will yield a positive on one side and a negative on the other. For zeros with odd multiplicity this alters the sign of our output and the function passes through the x-axis. Whereas the zero with even multiplicity will output a number with the same sign just above and below its zero, thus it 'bounces' off the x-axis.

The cubic function will increase more quickly than the quadratic, so the quadratic function must have a head start.  At , both functions evaluate to 8.  After than point, the cubic function will increase more quickly.

The domain was restricted to nonnegative values, so this interval is our only answer.

Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression fits the description of the zeroes.

Now we need to check the answer.

We are able to get back to the original expression, meaning that the answer is .

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is , which we can expland using the cube of a binomial pattern.

In the form given we can readily see that the polynomial has three zeros .  We observe that the zero at  has multiplicity of two whereas the others both have a multiplicity of one.  This make our answer .

The zeros of the polynomial are the values of  that set it equal to zero. In the given form we can clearly see that whenever any of the portions grouped in parentheses is equal to zero then the whole polynomial will be equal to zero. Since anything multiplied by zero is equal to zero. This fact lead us to our answer.

To find the zeros we need to factor this polynomial. We start by grouping the terms and pulling out a common factor.

now our two groups have a common factor so we pull that out and we get 

this can be factored further into 

and then we see that the x values that make each one of those terms 0 respectively are 

This is a 3rd order polynomial that can be factored.  The leading coefficient is 1, so only the second term in each factor need be considered.

, and so f factors into

.

At this point the shortcut

can be used.

Now the final factorization is

.

The zeros occur when each of the factors is zero, so

,

, and

.

For a polynomial to have 2 as a triple root, 2 must be it's solution 3 times. Solving this backwards, we start by constucting the final equation and working backwards.

The above equation gives us 2 as a triple root, so the polynomial on the left is what we are looking for.

Set y=0, then factor.  The leading coefficient is 12, so its factors need to be considered.  Trial and error will lead to 3 and 4 as the best choice combined with using 2 and -2 to give the last term.

Thus .

y=0 when either of the factors are 0.

So,

,

and

.