Solve the quartic equation x⁴ − 16x³ + 66x² − 16x − 35 = 0.

Step-by-step solution

1. 1

   Look for a repeated quadratic block. Half of $-16$ is $-8$, and $(x^2-8x)^2 = x^4 - 16x^3 + 64x^2$ already reproduces the top three terms up to a small remainder. That suggests the substitution

   $t = x^2 - 8x$

2. 2

   Rewrite the quartic in terms of t. Subtracting the expansion above:

   $x^4 - 16x^3 + 66x^2 - 16x - 35 = \left(x^2-8x\right)^2 + 2x^2 - 16x - 35 = t^2 + 2t - 35$

   The crucial coincidence is that the leftover $2x^2 - 16x$ is exactly $2t$, so no stray $x$ survives. A quartic in $x$ has become a quadratic in $t$.

3. 3

   Solve the quadratic in t. Two numbers with product $-35$ and sum $+2$ are $+7$ and $-5$:

   $t^2 + 2t - 35 = (t+7)(t-5) = 0 \quad \Longrightarrow \quad t = -7 \ \text{or} \ t = 5$

4. 4

   Translate back into two quadratics in x.

   $x^2 - 8x = -7 \ \Longrightarrow \ x^2 - 8x + 7 = 0$

   $x^2 - 8x = 5 \ \Longrightarrow \ x^2 - 8x - 5 = 0$

   Equivalently $x^4-16x^3+66x^2-16x-35 = \left(x^2-8x+7\right)\left(x^2-8x-5\right)$, which expands back to the original exactly.

5. 5

   Solve each quadratic. The first factors over the integers:

   $x^2 - 8x + 7 = (x-1)(x-7) = 0 \quad \Longrightarrow \quad x = 1 \ \text{or} \ x = 7$

   The second needs completing the square, since $\Delta = 64 + 20 = 84$ is not a perfect square:

   $(x-4)^2 = 21 \quad \Longrightarrow \quad x = 4 \pm \sqrt{21}$

6. 6

   Collect and check the four real roots.

   $x = 1,\quad x = 7,\quad x = 4 + \sqrt{21} \approx 8.5826,\quad x = 4 - \sqrt{21} \approx -0.5826$

   All four are real. Vieta's check: the roots sum to $1 + 7 + 8 = 16 = -(-16)/1$, and their product is $7 \cdot (16-21) = -35$, matching the constant term. Substituting $x = 4+\sqrt{21}$ numerically gives $0$ to twelve decimal places.