How to find total real roots of $x^4+4x^3+12x^2+7x-3$

Question

Question

The number of real values of x that satisfies the equation:

$$x^4+4x^3+12x^2+7x-3=0$$

Let $f(x)=x^4+4x^3+12x^2+7x-3$

My Approach

$f'(x)=4x^3+12x^2+24x+7$

$f''(x)=12x^2+24x+24$

$f''(x)=12(x^2+2x+2)$

Discriminant of $f''(x)$=$-4$

So, $f"(x)$ is always positive.

So, $f'(x)$ is increasing, and it has one or zero roots.

But, since it is cubic, it has 1 root.

Calculate $f(0)$= $-3$

Calculate $f(1)$= Positive

Calculate $f(-1) = -1$

Calculate $f(-2)= 15$

So, we know that there is a root between $0$ and $1$ and also between $-2$ and $-1$

Now, I'm stuck, How to prove that there are no more roots?

Any hints would help a lot.

Answer 1

Upon analyzing $f'(x)$,$f''(x)$,$f'''(x)$ [As you have done], we can conclude that $f'(x)$ only has one root, i.e, the graph of $f(x)$ will be similar to a parabola (slope initially is -ve then becomes +ve). So $f(x)$ will either have $0$ roots or $2$ roots.

$f(0)$ gives a negative value also $f(x)$ is similar to an upward parabola therefore it has $2$ roots.

Answer 2

As you have computed,

$f''(x) =12(x^2+2x+2) =12((x+1)^2+1) \gt 0$ and $f(x)$ has at least two roots.

If $f(x)$ has three or more roots, then $f'(x)$ has at least two roots so $f''(x)$ has at least one root which it does not.

Therefore $f(x)$ has exactly two roots.