Polynomials are Fun
Find the number of real solutions to: x 3 + x 2 − 6 x − 2 − x 1 = 0
The answer is 2.
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3 solutions
It is a fourth degree polynomial (of course after doing the common denominator). So the only candidates for the answer are 0,2 and 4, not 1 and 3 because if there were complex roots, they would occur in conjugates. So I tried 4, then 2 which turned out correct. :P
It would have been a nice multiple choice problem. You get only one try for that. :)
Without the aid of a graph. How would you know that the two functions intersected?
Here is the solution with brute force algebra.
Take x 3 + x 2 − 6 x − 2 − x 1 = 0 and multiply by x . We have a quartic equation x 4 + x 3 − 6 x 2 − 2 x − 1 = 0 . Now using the discriminant of a quartic evaluated at a = b = 1 we have Δ = c 2 d 2 − 4 c 3 d 2 − 4 d 3 + 1 8 c d 3 − 2 7 d 4 − 4 c 3 e + 1 6 c 4 e + 1 4 4 c d 2 e + 1 8 c d e − 8 0 c 2 d e − 6 d 2 e − 2 7 e 2 + 1 4 4 c e 2 − 1 2 8 c 2 e 2 − 1 9 2 d e 2 + 2 5 6 e 3 Now evaluating for our coefficients we have Δ = − 2 5 4 0 3 which indicates that the equation has two real roots and two complex roots. □
Just use a graphing calculator.
If we look at the − 2 − x 1 , it suggests that might be helpful to move this to the other side. So x 3 + x 2 − 6 x = 2 + x 1 . The left side factorises to give x ( x + 3 ) ( x − 2 ) so now we have a cubic with its roots, and a reciprocal, which can both be drawn on a graph relatively easily. Drawing the functions shows that there are two intersections, meaning that there are two real solutions.