Proving a parabola and polynomial intersect exactly 3 times

Question

the equations are : $P(x)=5-8(x-1)^2$ for the parabola and the other equation is $E(x)=(x-1)^5$ we need to prove that they intersect exactly 3 times.

What I did was compare them at first $5-8(x-1)^2 = (x-1)^5 $ and then make a new function $f(x)$ according to that we get $f(x)=(x-1)^5 -5+8(x-1)^2$and the function is continuous because its a polynomial and we can also have a derivative. I tried applying some numbers and i got $f(-1)=-5$ , $f(0)=2$ , $f(-5)=-7493$ , $f(2)=4$ so according to the Intermediate value theorem we have at least 3 roots but what we need to prove is that it has exactly 3 roots. and this looked like rolle's theorem

What I tried then was we want to show that it has exactly 3 roots according to Rolle's theorem so if the original equation has 4 roots the first derivative needs to have exactly 3

$f'(x)=5(x-1)^4 +16(x-1)$

And I got stuck here I could not figure a way to continue from here ..

Thanks for any tips and help all is appreciated and I would also like to know if anyone can confirm that what I did was right until the last step.

Sorry for my english mistakes hope it is understandable.

Answer 1

$$ Q = E-P = (x-1)^5 -5 + 8(x-1)^2$$

By descartes rule of signs, for the positive 'x' values this polynomial has a maximum of two real roots or no real roots.

Now shift $ x \to -x$

$$Q(-x) = -(x+1)^5 -5 + 8 (x+1)^2$$

By descartes rules of sign it has maximum of one roots for negative $x$ side

So, combining the two ideas, we get that it should have either one or three roots. Now, since you've identified existence of the three roots, this means the polynomial has only three real roots.

Answer 2

shifting the origin :$x-1=t$ we have to find the number of roots of $$f(t)=t^5-8t^2+5=0=(t-a)(t-b)(t-c)(t-d)(t-e)$$ here $a,b,c,d,e$ are its roots.Also one can evaluate $$a^2+b^2+c^2+d^2+e^2=0$$ which is not possible unless atleast one is complex or all are zero(which in this case is impossible) ; WLOG $e$ be complex then by $a+b+c+d+e=0$ we must have some other root must alos be complex hence $f(t)$ has minimum of of two complex roots, you have already found 3 real roots so...

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by comparing coefficients: $s=a+b+c+d+e=0$ $$w=ab+ac+ad+ae+bc+bd+be+cd+ce++de=0$$ also $$s^2-2w=a^2+b^2+c^2+d^2+e^2$$ hence $a^2+b^2+c^2+d^2+e^2=0$