I need someone to check my work. Thanks! This is a 2 mark homework question by the way. I am not sure why am I using such a long way to prove it. Is there a way to shorten it or is there a shorter, more intuitive method?
Proving that $x^3 +1=15x$ has at most three solutions. in the interval [-4,4].
Let $f(x)=x^3+1-15x$
Suppose for a contradiction, that this equation has at least 4 solutions, $a,b,c,d$, such that $f(a)=0,f(b)=0,f(c)=0,f(d)=0$. Since f is continuous and differentiable on $x\in\mathbb{R}$, by the Rolle's Theorem, there exist a $c_1 \in (a,b) , c_2 \in (b,c),c_3 \in (c,d) $, such that $f^\prime(c_1)=0,f^\prime(c_2)=0,f^\prime(c_3)=0 $
$f^\prime(x)=3x^2-15$
Moreover, if $f^\prime(x)$ has 3 solutions, by the Rolle's Theorem, Since f is continuous and differentiable on $x\in\mathbb{R}$, there exist a $d_1 \in (c_1,c_2) , d_2 \in (c_2,c_3)$, such that $f^{\prime\prime}(d_1)=0,f^{\prime\prime}(d_2)=0$
$f^{\prime\prime}(x)=6x$
Moreover, if $f^{\prime\prime}(x)$ has 2 solutions, by the Rolle's Theorem, Since f is continuous and differentiable on $x\in\mathbb{R}$, there exist a $e_1 \in (d_1,d_2)$, such that $f^{\prime\prime\prime}(e_1)=0$
$f^{\prime\prime\prime}(x)=6$
This implies that $f^{\prime\prime\prime}(e_1)=0=6$ Hence, we have a contradiction. Without loss of generality, we can apply the steps to cases where $f(x)=x^3+1-15x$ has 5 or more solutions and still achieve a contradiction. Therefore, the negation must be true, i.e $f(x)=x^3+1-15x$ has at most 3 solutions.
Best Answer
You only need to apply Rolle's Theorem once. A quadratic has at most two real solutions, this fact follows directly from the quadratic formula and doesn't need the Fundamental Theorem of Algebra. If you're willing to use the Fundamental Theorem of Algebra, then apply it directly to the cubic.
In response to OP's confusion.
The Fundamental Theorem of Algebra is a statement that a non-constant polynomial of degree $n$ has precisely $n$ roots (counting multiplicity) over the complex numbers. If we apply this theorem, then a cubic formula can have at most $3$ roots and we are in a direct contradiction if we have $4$. But as stated, the Fundamental Theorem of Algebra is overkill.
If you recall the quadratic formula, you will recall that if $x\in\mathbb{R}$ satisfies $$ax^2 + bx + c = 0$$ then $x$ must also satisfy $$x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$$ therefore we have (at most) only two possibilities for $x$ $$x=\frac{-b + \sqrt{b^2 - 4ac}}{2a}\ \ \ \text{or}\ \ \ x=\frac{-b-\sqrt{b^2 - 4ac}}{2a}$$ Therefore the quadratic formula is a direct proof that a degree two polynomial has at most $2$ distinct roots over the reals.