If a polynomial function has an axis of symmetry at x = c, is it true that c is equal to the average of the roots?

47 Views Asked by At

I was trying to learn more about cool substitutions to make my life easier when solving polynomial equations.

For example, in the problem $$x(x+1)(x+2)(x+3) = 24,$$ the LHS is symmetric at $x = -3/2$, which is the average of the zeroes, so we can do the substitution $$u = x - (-3/2)$$ and it would be easier to solve it that way.

I tried more examples like $$y = (x-5)(x-3)(x+5)(x+3)$$ and lo and behold, it is symmetric at $x = 0$.

Not every polynomial function has an axis of symmetry but if they do, is it always at $$x = \text{(avg. of roots)}?$$

1

There are 1 best solutions below

0
On BEST ANSWER

Yes. This is because if the function has axis of symmetry at $x=c$, then $p(c+x)=p(c-x)$. This means that for every root of the form $c+x_0$, you have another root of the form $c-x_0$.