Solve 8x^4 - 6x^3 + 30x^2 - 18x + 18 = 0.

Step-by-step solution

1. 1

   Remove the common numerical factor. Every coefficient is even:

   $8x^{4}-6x^{3}+30x^{2}-18x+18=2\left(4x^{4}-3x^{3}+15x^{2}-9x+9\right)$

   Since $2\neq 0$, the equation is equivalent to $4x^{4}-3x^{3}+15x^{2}-9x+9=0$. Smaller coefficients make the next step much easier.

2. 2

   Check that no rational root exists. Candidates are $\pm\frac{p}{q}$ with $p\mid 9$ and $q\mid 4$. All coefficients alternate in a way that makes every positive candidate give a positive value, and every negative candidate gives a positive value too, so there is no linear factor over the rationals. The next hope is a factorisation into two quadratics.

3. 3

   Set up undetermined coefficients. Try

   $4x^{4}-3x^{3}+15x^{2}-9x+9=\left(4x^{2}+bx+3\right)\left(x^{2}+ex+3\right)$

   choosing the constant terms as $3$ and $3$ because their product must be $9$. Expanding gives

   $4x^{4}+(4e+b)x^{3}+(be+15)x^{2}+3(b+e)x+9$

4. 4

   Match coefficients and solve the small system. Comparing with the target:

   $4e+b=-3,\qquad be+15=15,\qquad 3(b+e)=-9$

   The middle equation forces $be=0$, so $b=0$ or $e=0$. With $e=0$ the third gives $b=-3$, and the first checks out: $4(0)+(-3)=-3$ ✓. Hence

   $4x^{4}-3x^{3}+15x^{2}-9x+9=\left(4x^{2}-3x+3\right)\left(x^{2}+3\right)$

5. 5

   Solve the first quadratic. For $4x^{2}-3x+3=0$ the discriminant is

   $(-3)^{2}-4(4)(3)=9-48=-39<0$

   so

   $x=\frac{3\pm i\sqrt{39}}{8}$

6. 6

   Solve the second quadratic and conclude. $x^{2}+3=0$ gives $x=\pm i\sqrt3$. All four roots are non-real, so the quartic has no real solutions — consistent with the graph, which stays above the axis (its minimum value is positive). Multiplying the two factors back out and doubling reproduces $8x^4-6x^3+30x^2-18x+18$ exactly ✓.