Solve the quartic equation u⁴ − 6u³ + 8u² + 6u − 9 = 0.

Step-by-step solution

1. 1

   Set up an unknown split into two quadratics. Rather than hunting one root at a time, assume the quartic breaks into two monic quadratics with integer coefficients:

   $u^4 - 6u^3 + 8u^2 + 6u - 9 = \left(u^2 + au + b\right)\left(u^2 + cu + d\right)$

   This is worth trying first because a successful split hands you all four roots at once.

2. 2

   Expand and match coefficients. Multiplying out the right side:

   $u^4 + (a+c)u^3 + (ac+b+d)u^2 + (ad+bc)u + bd$

   so the four unknowns satisfy

   $a + c = -6, \quad ac + b + d = 8, \quad ad + bc = 6, \quad bd = -9$

3. 3

   Search the small integer possibilities for b and d. Since $bd = -9$, the candidate pairs are $(\pm1, \mp9)$, $(\pm3, \mp3)$, $(\pm9, \mp1)$. Trying $b = -1$, $d = 9$ makes the second equation collapse:

   $ac - 1 + 9 = 8 \quad \Longrightarrow \quad ac = 0$

   So one of $a, c$ is zero — a big simplification, and the reason this pair is the one to test first.

4. 4

   Finish the coefficient solve and check all four equations. With $ac = 0$ take $a = 0$, hence $c = -6$. Then

   $ad + bc = 0 \cdot 9 + (-1)(-6) = 6 \ \checkmark$

   All four conditions hold, so

   $u^4 - 6u^3 + 8u^2 + 6u - 9 = \left(u^2 - 1\right)\left(u^2 - 6u + 9\right)$

   Expanding back gives $u^4 - 6u^3 + 9u^2 - u^2 + 6u - 9$, which is the original polynomial.

5. 5

   Factor each quadratic completely. One is a difference of squares and the other a perfect square:

   $u^2 - 1 = (u-1)(u+1), \qquad u^2 - 6u + 9 = (u-3)^2$

   $\Longrightarrow \quad (u-1)(u+1)(u-3)^2 = 0$

6. 6

   Read off the roots and their multiplicities.

   $u = 1, \qquad u = -1, \qquad u = 3 \ \text{(double root)}$

   Substituting confirms each: $u=1$ gives $1-6+8+6-9=0$, $u=-1$ gives $1+6+8-6-9=0$, and $u=3$ gives $81-162+72+18-9=0$. Because $u = 3$ has multiplicity two, the graph touches the $u$-axis there without crossing, and the four roots counted with multiplicity sum to $1 - 1 + 3 + 3 = 6$, matching $-(-6)/1$.