Solve x^6 + 4x^5 - 4x^4 + 4x^3 - 7x^2 - 8x + 10 = 0.

Step-by-step solution

1. 1

   Hunt for a rational root first. By the rational root theorem, any rational root divides the constant $10$: the candidates are $\pm1,\pm2,\pm5,\pm10$. Testing $x=1$ is quickest because it just sums the coefficients:

   $1+4-4+4-7-8+10=0$

   So $x=1$ is a root and $(x-1)$ is a factor. A degree-six equation is otherwise hopeless by formula, so finding one root is the essential first break.

2. 2

   Divide out $(x-1)$ by synthetic division. With the coefficients $1,4,-4,4,-7,-8,10$ and divisor value $1$, the quotient is

   $x^{5}+5x^{4}+x^{3}+5x^{2}-2x-10$

   and the remainder is $0$, confirming the root.

3. 3

   Factor the quintic by grouping. The coefficients pair up as $(1,5),(1,5),(-2,-10)$ — each pair in the ratio $1:5$, which is the signature of a common factor $(x+5)$:

   $x^{5}+5x^{4}+x^{3}+5x^{2}-2x-10=x^{4}(x+5)+x^{2}(x+5)-2(x+5)=(x+5)\left(x^{4}+x^{2}-2\right)$

4. 4

   Treat the quartic as a quadratic in $x^{2}$. With $y=x^{2}$,

   $y^{2}+y-2=(y+2)(y-1)\ \Longrightarrow\ x^{4}+x^{2}-2=\left(x^{2}+2\right)\left(x^{2}-1\right)=\left(x^{2}+2\right)(x-1)(x+1)$

   Note the second appearance of $(x-1)$ — that is where the double root comes from.

5. 5

   Assemble the complete factorisation.

   $(x-1)^{2}(x+1)(x+5)\left(x^{2}+2\right)=0$

   Expanding this back reproduces the original coefficients $1,4,-4,4,-7,-8,10$ exactly ✓, which is the check that the whole chain of steps held together.

6. 6

   Read off the six roots. Setting each factor to zero:

   $x=1\ \text{(double)},\quad x=-1,\quad x=-5,\quad x=\pm i\sqrt2$

   Counted with multiplicity these are six roots, as the fundamental theorem of algebra requires for a degree-six polynomial. Four are real (with $1$ repeated) and two are a conjugate imaginary pair.