Question

The number of distinct real roots of \[{x^4} - 4{x^3} + 12{x^2} + x - 1 = 0\] is
A) 2
B) 3
C) 4
D) 5

Answer 1

Hint: You can find the distinct real roots for the equation by number of methods.
1) Descartes rule of Signs
2) Factorization and simplification
3) By differentiating the polynomial

Complete step-by-step answer:
Given:
\[{x^4} - 4{x^3} + 12{x^2} + x - 1 = 0\]
Let:
\[f(x) = {x^4} - 4{x^3} + 12{x^2} + x - 1 = 0\]
Note that the pattern of signs of the coefficients is +1, −4, +12, +1, −1.
Our variables goes from positive (1) to negative (-4) to positive (12) to positive (1) to negative (-1).
Between the first two coefficients there are first change in signs and between our second and third we have our second change, then between our third and fourth we have no change in signs and between our 4th and 5th coefficients we have a third change of coefficients.
Descartes´ rule of signs tells us that we then have exactly 3 real positive zeros or less but an odd number of zeros. Hence our number of positive zeros must then be either 3, or 1.
Also note that:
\[{x^4} - 4{x^3} + 12{x^2} + x - 1 = 0\]
\[ \Rightarrow {x^4} - 4{x^3} + 6{x^2} - 4x + 1 + 6{x^2} - 12x - 6 + 17x - 17 - 9 = 0\]
\[ \Rightarrow {(x - 1)^4} + 6{(x - 1)^2} + 17(x - 1) + 9 = 0\]
Note that the coefficients of all the expressions in \[(x - 1)\;\]on the right hand side are positive.
So there is no\[x \geqslant 1\;\]for which this is zero.
Hence we can deduce that any real zeros of \[{x^4} - 4{x^3} + 12{x^2} + x - 1 = 0\]are in \[(0,1)\].
\[
   \Rightarrow f'(x) = 4{x^3} - 12{x^2} + 24x + 1 \\
   \Rightarrow f'(x) = 4({x^3} - 3{x^2} + 3x - 1) + 12(x - 1) + 17 \\
   \Rightarrow f'(x) = 4{(x - 1)^3} + 12(x - 1) + 17 \\
 \]
which is positive for all \[x \in (0,1)\]
Hence \[f(x)\;\] can only have one zero in \[(0,1)\], making a total of 2 real zeros including the negative one.

So, option (A) is the correct answer.

Note: Descartes' rule of sign is used to determine the number of real zeros of a polynomial function.
It tells us that the number of positive real zeros in a polynomial function f(x) is the same or less than by an even number as the number of changes in the sign of the coefficients. The number of negative real zeros of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.
Some of the points of Descartes Rule:-
If the coefficients of \[f(x)\;\]are all positive then the equation has no positive roots.
If the coefficients of even powers of x are all of one sign and the coefficients of odd powers of x are all of opposite signs then the equation has no negative root.
If the equation contains only even powers of x and the coefficients are of the same sign, then the equation will have no real root.
If the equation contains only odd powers of x and the coefficients are of the same sign, then the equation will only have 0 real root.