(Becuase there is a "pre-calculus" tag, I will use calculus as little as possible in this argument for the original problem, since a couple of points can only be sufficiently refined by applying the derivative function.)
We might start with the function $ \ h(x) \ = \ x^4 - x^3 \ \ , \ $ which has a triple zero at the origin and its fourth (real) zero at $ \ x \ = \ 1 \ \ ; $ because $ \ h(x) \ > \ 0 \ $ for $ \ x \ < \ 0 \ $ and $ \ x \ > \ 1 \ \ , \ $ we must have $ \ h(x) \ < \ 0 \ $ in the interval $ \ (0 \ , \ 1) \ \ . \ $ We find $ \ \mathbf{h \left( -\frac12 \right) \ = \ \frac{3}{16} } \ \ , \ \ h \left( -\frac14 \right) \ = \ \frac{5}{256} \ \ , \ \ h \left( \frac14 \right) \ = \ \frac{3}{256} \ \ , \ \ h \left( \frac12 \right) \ = \ -\frac{1}{16} \ \ , $ $ h \left( \frac34 \right) \ = \ -\frac{27}{256} \ \ . \ $ (This last is in fact the global minimum of the function.) So the "bottom" of the function curve is rather "flat" with a modest "dip" in the interval $ \ \left( \frac12 \ , \ 1 \right) \ \ . $
Modifying this to $ \ g(x) \ = \ x^4 - x^3 + ax \ \ , \ $ has the effect of "tilting the floor" of the function curve. We will want to lower the portion of the curve "to the left" of the $ \ y-$axis while raising the portion "to the right" by taking $ \ a \ > \ 0 \ : \ $ this will create a "turning point" to the left of the origin and also to its right, since, in the neighborhood of the origin, $ \ g(x) \ $ is very close to just $ \ ax \ \ . $ At the $ \ x-$coordinates listed above, the function values are now
$$ \ \mathbf{g \left( -\frac12 \right) \ = \ \frac{3}{16} - \frac{a}{2} } \ \ , \ \ g \left( -\frac14 \right) \ = \ \frac{5}{256} - \frac{a}{4} \ \ , \ \ g \left( \frac14 \right) \ = \ \frac{3}{256} + \frac{a}{4} \ \ , $$ $$ g \left( \frac12 \right) \ = \ -\frac{1}{16} + \frac{a}{2} \ \ , \ \ g \left( \frac34 \right) \ = \ -\frac{27}{256} + \frac{3a}{4} \ \ . \ $$
We still have a zero at $ \ x \ = \ 0 \ \ , \ $ but it is now possible to have zeroes in $ \ \left( -\frac12 \ , \ -\frac14 \right) \ $ and $ \ \left( \frac14 \ , \ \frac34 \right) \ \ . $ The zero at $ \ x \ = \ 1 \ $ of $ \ h(x) \ $ "moves" to $ \ x \ < \ 1 \ $ since $ \ g(1) \ = \ a \ \ . $
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EDIT (8/6) -- If we bring in differentiation to investigate the number of turning points of the function, we have $ \ g'(x) \ = \ 4x^3 - 3x^2 + a \ \ , \ $ which yields the "depressed" cubic polynomial $ \ \mathcal{G}(x) \ = \ t^3 \ - \ \frac{3}{16}·t \ + \ \frac14·\left(a - \frac18 \right) \ \ . $ In order for $ \ g(x) \ $ to have three turning points, the discriminant must be
$$ \Delta \ \ = \ \ 4·\left( -\frac{3}{16} \right)^3 \ + \ \frac{27}{4^2}·\left(a - \frac18 \right)^2 \ \ = \ \ \frac{27}{64}·a·\left(a - \frac14 \right) \ \ < \ \ 0 \ \ , $$
which is true in the interval $ \ 0 \ < \ a \ < \ \frac14 \ \ $ (in agreement with comments to the original post).
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We see, however, that there is a further limit to the magnitude of $ \ a \ $ in order to avoid raising the turning point near $ \ x \ = \ \frac34 \ $ above the $ \ x-$axis. So we must have $ \ -\frac{27}{256} + \frac{3a}{4} \ \le \ 0 $ $ \Rightarrow \ 0 \ \le \ a \ \le \ \sim \frac{9}{64} \ \approx \ 0.141 \ . \ $ (A more precise value turns out to be $ \ 0.148 \ \ . \ ) $ This gives us $ \ \frac{3}{16} \ \ge \ g \left( -\frac12 \right) \ \ge \ \sim \frac{3}{16} - \frac{9/64}{2} \ \ . $
We can then make a "vertical shift" by amending the function expression to $ \ f(x) \ = \ x^4 - x^3 + ax + b \ \ . \ $ We note that for $ \ a \ > \ 0 \ \ , \ \ b \ < \ 0 \ \ $ that the Rule of Signs will admit the possibility of as many as three positive real zeroes and one negative one; with $ \ b \ > \ 0 \ \ , \ $ we may have two of each. (Having $ \ a \ < \ 0 \ $ only allows a total of two "sign changes" for $ \ f(x) \ $ and $ \ f(-x) \ \ , \ $ which is consistent with our result from examining the derivative function.)
However, the turning points are rather "shallow", so $ \ | \ b \ | \ $ must be fairly small; otherwise, two of the consecutive turning points will be on the same side of the $ \ x-$axis, which will leave us with only two intercepts. For the purpose of answering the original question, we do not need to consider the effect of lowering the curve $ \ ( \ b \ < \ 0 \ ) \ \ , \ $ since we already have $ \ g \left( -\frac12 \right) \ = \ \frac{3}{16} - \frac{a}{2} \ \le \ \frac{3}{16} \ \ . \ $ The remaining question is whether raising the curve $ \ ( \ b \ > \ 0 \ ) \ $ can overcome the amount by which $ \ h \left( -\frac12 \right) \ $ has been reduced.
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EDIT (8/6) [replacing the "pre-calculus" argument] -- The complication in finding the limit of the amount of "vertical shift" that will still permit $ \ f(x) \ $ to have four $ \ x-$intercepts (four real zeroes) is that the turning points are asymmetrical; we would really prefer not to have to locate the relative minima of $ \ f(x) \ $ and evaluate them to find the limit on $ \ b \ \ . $ Instead, we will construct a symmetrical function by using the value of $ \ a \ $ that is in the center of the permissible interval, $ \ \mathfrak{G}(x) \ = \ x^4 - x^3 + \frac18·x \ \ . \ $ It is symmetrical about $ \ x \ = \ \frac14 \ \ , \ $ producing
$$ \left(u + \frac14 \right)^4 \ - \ \left(u + \frac14 \right)^3 \ + \ \frac18·\left(u + \frac14 \right) \ \ = \ \ u^4 \ - \ \frac38·u^2 \ + \ \frac{5}{256} \ \ , $$
which factors as $ \ \left(u^2 - \frac{1}{16} \right)·\left(u^2 - \frac{5}{16} \right) \ \ , $ so there are indeed four real zeroes. The extrema are found from
$$ \ \frac{d}{du} \ \left[ \ \left(u^2 - \frac{1}{16} \right)·\left(u^2 - \frac{5}{16} \right) \ \right] \ \ = \ \ 4·u·\left(u^2 - \frac{3}{16} \right) \ \ = \ \ 0 \ \ . $$
Hence, there is a maximum at $ \ u \ = \ 0 \ $ and two minima at $ \ u \ = \ \pm \frac{\sqrt3}{4} \ \ $ at which the minimum value for
$ \ \mathfrak{G}(x) \ $ is given as $ \ \left( \pm \frac{\sqrt3}{4} \right)^4 - \frac38·\left( \pm \frac{\sqrt3}{4} \right)^2 + \frac{5}{256} \ = \ -\frac{4}{256} \ = \ -\frac{1}{64} \ \ . $
So $ \ \mathfrak{G}(x) \ + \ \frac{1}{64} $ is raised just to the point where its minima are tangent to the $ \ x-$axis, that is, at which its four distinct real zeroes become two real double zeroes. (By a similar argument, we cannot "lower" $ \ \mathfrak{G}(x) \ $ by more than the "height" of its maximum at $ \ x \ = \ \frac14 \ \ , \ $ which is $ \ \frac{5}{256} \ \ . \ $ Thus, for all four zeroes of $ \ f(x) \ $ to be real, we have at most $ \ 0 \ < \ a \ < \ \approx 0.148 \ $ and $ \ -\frac{5}{256} \ \le \ b \ \le \ \frac{1}{64} \ \ . \ $ (As we've said, the range for $ \ b \ $ is often much narrower.)
The function value at $ \ x \ = \ -\frac12 \ $ is then $ \ \large{ \mathbf{f \left( -\frac12 \right) \ \le \ \frac{3}{16} - \frac{a}{2} + b } } \ \ , \ $ with (generally) $ \ | \ b \ | \ \le \ \frac{a}{8} \ \ . \ $
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[For illustrative purposes, I kept my "pre-calculus" argument here; it turns out that it is fairly accurate:
With $ \ 0 \ < \ a \ < \ \frac{9}{64} \ \ $ and $ \ b \ > \ 0 \ \ , \ $ we cannot then "raise" the curve much more than to place $ \ f \left( -\frac18 \right) \ = \ \frac{9}{4096} - \frac{a}{8} + b \ = \ 0 \ \ , \ $ which is roughly where a turning point is located; this gives us $ \ b \ \approx \ \frac{a}{8} \ \ , \ $ so $ \ 0 \ < \ b \ < \ \frac{9}{512} \ \ . \ $ The function value for $ \ f(x) \ $ at $ \ x \ = \ -\frac12 \ \ $ is then no larger than about
$$ f \left( -\frac12 \right) \ \ = \ \ \frac{3}{16} \ - \ \frac{9/64}{2} \ + \ \frac{9}{512} \ < \ \frac{3}{16} \ \ . \ ] $$