Given polynomial

$f(x)=1+2x+3x^2+4x^3=4x^3+3x^2+2x+1.$

We need the sum of all distinct real roots of $f(x)$.

Factor the polynomial

Group terms:

$4x^3+3x^2+2x+1=(4x^3+2x)+(3x^2+1).$

This is not immediately helpful. Instead, try factoring as:

$4x^3+3x^2+2x+1=(x^2+1)(4x+1)-x^2(4x+1).$

Better yet, factor directly:

$4x^3+3x^2+2x+1=(x+1)(4x^2-x+1).$

Check:

$(x+1)(4x^2-x+1)=4x^3-x^2+x+4x^2-x+1=4x^3+3x^2+2x+1.$

So,

$f(x)=(x+1)(4x^2-x+1).$

Find real roots

From $x+1=0$, we get

$x=-1.$

Now check the quadratic:

$4x^2-x+1=0.$

Its discriminant is

$\Delta = (-1)^2-4(4)(1)=1-16=-15<0.$

So this quadratic has no real roots.

Hence the only real root of $f(x)$ is

$x=-1.$

Compute $s$ and $t$

Since there is only one distinct real root,

$s=-1.$

Therefore,

$t=|s|=|-1|=1.$

Locate the interval containing $t$

We have

$t=1.$

Check options:

• A: $\left(-\frac14,0\right)$ → does not contain $1$
• B: $\left(-11,-\frac34\right)$ → does not contain $1$
• C: $\left(-\frac34,-\frac12\right)$ → does not contain $1$
• D: $\left(0,\frac14\right)$ → does not contain $1$

So none of the given intervals contains $t=1$.

Comparison with stored answer

Stored correct answer is C, but option C is a negative interval and cannot contain $t=|s|\ge 0$. Since $t=1$, the stored answer is inconsistent with the question.

It is likely that either:

• the question intended to ask for the interval containing $s$ instead of $t$, in which case $s=-1$ would lie in option B, or
• the options were misprinted.

As written, there is no correct option among A, B, C, D.