Determining the number of non-real roots. Multiple choice strategy.

I was tutoring a student, and this question came up on a mock standardized test. Assume the student has learned advanced precalculus at a high school level and has access to a graphing calculator. The question is meant to be solved in a few minutes, tops.

Which of the following equations has exactly two non-real solutions?

(A) $$\quad x^3-x^2+2x-2 = 0$$

(B) $$\quad x^3 -2x^2 -5x +6 = 0$$

(C) $$\quad x^4-7x^2+12=0$$

(D) $$\quad x^3-8x^2+11x+20=0$$

(E) $$\quad x^4-5x^3+x^2 +25x-30=0$$

The first strategy that came to my mind is using Descartes's Rule of Signs, but that test seems to be inconclusive. A graphical approach seems to be another way to approach it, but that seems inefficient to me, and not entirely reliable in a test taking scenario- where you may fumble with manipulating window sizes and whatnot.

The answer is cited as (A), and indeed, the first answer choice can be easily factored as $$(x^2+2)(x-1)$$, which verifies it as a solution (note: I haven't checked the roots of the other polynomials). But again, I do not expect that factoring should have been the intended approach here.

Does anyone see a quick method to solve this, or is this a poorly designed question?

Edit: To followup on this question. Indeed, Gregory was correct in the comments. This question was from a unofficial SAT II Math practice exam. While the SAT II Math test does not require a calculator for any of its questions, it doesn't prohibit the use of one. It doesn't even ban computer algebra systems, like those on, say, the TI-89. Now, this particular practice test was constructed with (some) questions designed to train students to use the algebra software.

Thanks for the suggested solutions, though. I like the calculus methods, and I think my student will appreciate them as well.

• If the person has a calculator, couldn't you just plug each one in and see how many roots you have? – Gregory Apr 14 '17 at 20:08
• @Gregory You're probably right. Yeah, this test doesn't explicitly ban computer algebra systems. I bet this mock test was training the use of one... How lame. The real test doesn't require such tools, which is why I tend not to train students to rely on their calculators. – suneater Apr 14 '17 at 20:12
• Take the factors of the first and last number and use synthetic subtitution, find the real zeros. From there use the quadratic formula using the numbers made in the equation with the zero and find the non real solutions. It might take a while, but it is one way to do it. – user242559 Apr 14 '17 at 20:13
• It looks like (A) and (C) have easy factorizations: $(x-1)(x^2 + 2)$ and $(x^2 - 3)(x^2 - 4)$. Thus (A) is one of the answers. (B) has $x-1$ and (D) has $x+1$ as a factor (seen by inspection) and can be factored completely using synthetic division. The last one might require a bunch of guess and check. – Umberto P. Apr 14 '17 at 20:15
• I'm curious to see this short and efficient method. – Gregory Apr 14 '17 at 20:16

If your test-taking strategy is savvy, you will try to eliminate the easier possibilities first; here, that means looking at the cubics first. Advanced pre-calc nowadays mentions how to take the derivative of a polynomial, so you can look at (A) and find that it has no real minimum or maximum, because the derivative $$3x^2-2x+2$$ has a discriminant of $b^2-4ac = 4-24 < 0$. (That is, in trying to solve for $x$ using the quaadratic formula, you get no real solutions.) So (A) is monotone increasing, and can have only one point at which it crosses $y=0$, thus exactly two non-real solutions.

FInally, using the fact that this is a test question with a unique correct choice, you know the others all have zero (or 4) non-real solutions.

• You mean $3x^2$ I suppose. – user58697 Apr 14 '17 at 21:29

Hints for the quartics:

• (C) $\;x^4-7x^2+12=0$  This is a biquadratic i.e. a quadratic in $x^2\,$. By the discriminant and rule of signs it's easy to tell that it has two positive roots in $x^2\,$, therefore $4$ real roots in $x$.

• (E) $\;x^4-5x^3+x^2 +25x-30=x^4-5x^3+6x^2 - 5\cdot (x^2-5x+6)\,$.

• Thanks for the factorization of (E). – suneater Apr 14 '17 at 20:45

If this were a MCQ with one or more than one correct answers, We'll need to check every option (You haven't specified explicitly that's why I'm taking full proof case)

Let $f$ be your cubic function. Just take derivative of each function. By the roots of $f'$ (Which is obviously a quadratic in case of cubic .. ). You will get the point of local maxima and minima of $f$.Let them be at $x=\alpha$ and $x=\beta$.

Now all you have to do is check sign of $f(\alpha)$ and $f(\beta)$ (Suppose $\alpha > \beta$ and leading coefficient to be positive.)

There are $4$ possibilities :

• $f(\alpha) >0$ ,$f(\beta) >0 \implies$ the function turns above $x$-axis and thus only one real root, rest two are complex.

• $f(\alpha) > 0$ ,$f(\beta) <0 \implies$ the function turns above $x$-axis and again below $x$-axis and thus only three real roots.

• $f(\alpha) <0$ ,$f(\beta) <0 \implies$ the function turns below $x$-axis and both the times thus only one real root, rest two are complex..

• $f(\alpha) <0$ ,$f(\beta) >0~$ Never possible.

• If $f'$ doesn't has real roots $\implies$ the cubic is monotonously increasing .Hence one real (repeated) root.

This method works well for cubic polynomials.

For bi-quadratic given in option $(C)$, it's actually a quadratic in $x^2$.

For biquadratic in option $(E)$, We'll need to factorize it as $(x^2-5x+6)(x^2-5)$ (as done by @dxiv in his answer).

Rest are all cubics.

• The question is tagged precalculus. – Umberto P. Apr 14 '17 at 20:43
• @UmbertoP. Than it seems to be impossible to do in few minutes. – Jaideep Khare Apr 14 '17 at 20:52