Sixth degree polynomial

Consider $x^6+ax^5+bx^4+x^3+bx^2+ax+1=0$ , one of whose roots is $x=1$ . First, use synthetic division to divide out the root $x=1$ . This leads to $x^5+(a+1)x^4+(a+b+1)x^3+(a+b+2)x^2+(a+2b+2)x+2a+2b+2=0$ , along with the fact that $2a+2b+3=0$ (because the remainder must equal zero). Next, attempt divide out the root again. The next remainder is $6a+6b+9$ , which must equal zero by the previous condition. Therefore, $x=1$ is a double root. Then substitute $-\frac{3}{2}-a$ for $b$ to obtain $x^4+(a+2)x^3+(a+\frac{3}{2})x^2+(a+2)x+1=0$ . Divide the equation by $x^2$ and then rearrange it to the following form: $\Big(x+\frac{1}{x}\Big)^2+(a+2)\Big(x+\frac{1}{x}\Big)+a-\frac{1}{2}=0$ . By the quadratic formula, $x+\frac{1}{x}=\frac{-a-2\pm\sqrt{a^2+6}}{2}$ . The equation $x+\frac{1}{x}=y$ has a real solution in $x$ iff $y\notin (-2,2)$ . For the equation $x+\frac{1}{x}=\frac{-a-2-\sqrt{a^2+6}}{2}$ , the root(s) are real iff $a\geq -\frac{1}{2}$ . And for the equation $x+\frac{1}{x}=\frac{-a-2+\sqrt{a^2+6}}{2}$ , the root(s) are real iff $a\leq -\frac{5}{2}$ . Therefore, at least one of the equations must have imaginary solutions. Thus, an upper bound on the number of distinct real roots (of the equation $x^6+ax^5+bx^4+x^3+bx^2+ax+1=0$ , given that $x=1$ is a root) is $3$ . This upper bound is achievable for many values of $a$ . As a specific example, take $a=0$ . The fact that the equation has two distinct real roots (after the double root $x=1$ has been eliminated) is easily checked by taking $y=\frac{-a-2-\sqrt{a^2+6}}{2}$ and $a=0$ , and then examining the discriminant of $x^2+yx+1=0$ . One gets $y^2-4=\Big(\frac{-2-\sqrt{6}}{2}\Big)^2-4=\sqrt{6}-2>0$ . (It's also obvious that $x=1$ is not a root of the equation $x^4+2x^3+\frac{3}{2}x^2+2x+1=0$ .)

We know p(1) = 0.

i.e. 1 + a.1 + b.1 + 1 + b.1 + a.1 + 1 = 2a + 2b +3 = 0 ..(i)

The above has to be always true except for ordered pair which satisfy x = -1.

Now, p'(x) = 6x $^{5}$ + 5ax $^{4}$ + 4bx $^{3}$ + 3x $^{2}$ + 2bx + a

p'(1) = 6 + 5a +4b +3 +3b +a = 3(2a + 2b +3)

Now, from (i) we can say that p'(1) is always zero. Hence, x=1 is a double root. Therefore, a sixth-degree polynomial has 6 maximum roots but the question has asked for distinct roots so the answer is 5.