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In my textbook, I was given the problem:

Finds the roots of the equation $3x^4 + 4x^3 + x - 1 = 0$
a. three positive real numbers and one negative real number
b. three negative real numbers and one positive real number
c. one negative real number and three imaginary numbers
d. one positive real number, one negative real number, and two imaginary numbers
e. two positive real numbers, one negative real numbers, and one imaginary number

By looking at the graph of the function, you see that it passes through the x-axis on both the positive and negative sides, so you know that it must have a positive and negative real root. Given the choices here, you can then deduce that the answer (which is correct) is

d. one positive real number, one negative real number, and two imaginary numbers

simply because if imaginary roots must come in conjugate pairs if there are no imaginary coefficients in the simplified form of the equation. However, the textbook's reasoning confuses me:

Observe that the graph crosses the x-axis twice - once at a positive x value and once at a negative one. Since the function is a degree 4 polynomial, there are 4 roots, so the other two must be complex conjugates.

The explanation makes sense for the most part, but I am confused as to why one would be able to discredit the function from having 2 pairs of double real roots. This is not one of the options to this question, so this isn't the case here, but I'm curious if there is a way to tell if something has double real roots or not (in a timed environment, such as a test, because I am already aware of one of the long ways to do this: factoring and using synthetic division to find all the roots).

I tried this out with the equation $(x-2)^4$. I was thinking that if you wanted to test if 2 was a root, you would plug it in $(2-2)^4$ and see the result is 0, so it is a factor (according to the factor theorem). I attempted to continue this logic and figured that in $(x-2)^4$, 2 is a factor 4 times, so I attempted to plug in $((2^4)-2)^4$. However, off the bat I could see this idea wouldn't work. I am wondering if there is a quick way to tell if something has double, quadruple, etc. real roots or if something has complex conjugate pairs of imaginary roots.

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  • $\begingroup$ this a polymial equation of degree four and solvable by radicals i would plot the function $$f(x)=3x^4+4x^3+x-1$$ and after this try to prove the Statements above $\endgroup$ – Dr. Sonnhard Graubner Jan 7 '17 at 18:33
  • $\begingroup$ If you compute the derivative and take the $\gcd$ you can verify that it has no multiple roots. Is that a technique you are comfortable with? $\endgroup$ – lulu Jan 7 '17 at 18:34
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    $\begingroup$ Informally, you can see from the graph that the tangent isn't horizontal near the roots...which is hardly rigorous but which might be persuasive. To illustrate the point, look at the graph of $x^3(x-1)$. That's another quartic that crosses the real line exactly twice...but here the graph looks perfectly flat near $0$. $\endgroup$ – lulu Jan 7 '17 at 18:36
  • $\begingroup$ If there's multiple roots, then $f(x)$ is either $(x-a)^4$ or $(x-b)^3(x-c)$ or $(x-d)^2(x-e)^2$ or $(x-f)^2(x-g)(x-h)$ or $(x-i)^2(x^2+jx+k)$. Without calculus, you still can look at the root whether it's local max/min or point of inflection. $\endgroup$ – Ng Chung Tak Jan 7 '17 at 18:37
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    $\begingroup$ Furthermore, it's not easy to determine whether $f(x)$ is in the form of $(x-a)^4$ or $(x-i)^2(x^2+jx+k)$, unless you use derivative test (that requires calculus). $\endgroup$ – Ng Chung Tak Jan 7 '17 at 18:45
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If a polynomial $f$ has a double root at $a$, then $f'(a) = 0$ as well. You can see in this case that $f'(x) > 0$ for $x\ge 0$, so $f$ has only one non-repeated positive root. If $f$ has no complex roots, then that negative root must have been repeated thrice, so $f$ must have the form $c(x-a)^3(x-b)$, with $a<0$ and $b>0$. Expanding and comparing the terms with $x^2$, we get $a=0$ or $b=-a$. Both of which are impossible.

Therefore, $f$ has at least one complex root. But for polynomials with real coefficients, complex roots come in conjugates. Hence the conclusion.

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For large positive values of $x$, $f(x) = 3x^4 + 4x^3 + x - 1$ is always positive. Since $f(x)$ is a polynomial of degree $4$, the curve opens upwards and hence $f(x)$ must also be positive for large negative values of $x$. However, $f(0) = -1 < 0$. Hence $f(x)$ must have at least $2$ real roots.

Combine this with @Open Ball's reasoning and you have your answer as option d.

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