Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0)$ and $(-2,0)$ and $y$ intercept of $(0,18)$. Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0) $ and $ (-2,0)$ and $y$ intercept of $(0,18)$.
The root ($\frac{1}{2},0)$ has multiplicity $2$.
I am to write the factored form of the polynomial with the above information. I get:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Whereas the provided solution is:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$
Here's my working:
Write out in factored form:
$f(x) = a\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
I know that $f(0)=18$ so:
$$18 = a\big(-\frac{1}{2}\big)^2(2)(-6)$$
$$18 = a\big(\frac{1}{4}\big)(2)(-6)$$
$$18 = -3a$$
$$a = -6$$
Thus my answer:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Where did I go wrong and how can I arrive at:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$ ?

You didn't go wrong anywhere.
$-6(x-\frac{1}{2})^2(x+2)(x-6) = -\frac{3}{2}(2x-1)^2(x+2)(x-6)$
That is because
$-6(x- \frac 1{2})^2 = -6(\frac 12[2x-1])^2= -6\cdot (\frac 12)^2(2x-1)^2 = -6\cdot \frac 14(2x-1) = -\frac 32(2x-1)$.
Is there some rule that that says fractions in the $(ax + b)$ parts are frowned upon?
If so, if you get $(x + \frac ab)^k$ and can replace it with $(\frac 1b)^k(bx + a)^k$ but I don't see why you should have to.
(In fact, I much prefer your notation as it directly indicates the roots and solutions... and indicates what the leading coeficient will actually be when expanded out.  And what the heck kind of sense does removing a horrifying offensive fraction from one area make if you are just going to have to put an equally offensive fraction somewhere else?)
But.... weren't you supposed to expand this out? So far as I can tell neither answer has done that.  If you expand it out, you will see both answers are exactly the same.
