1. **Problem Statement:** We are given a quartic equation in $x$: $$x^4 + b x^3 + c x^2 + d x + e = 0$$ and a substitution $x = y - \frac{b}{4a}$ (with $a=1$ here) to remove the cubic term and simplify the equation. 2. **Goal:** Rewrite the quartic in terms of $y$ and express it in a depressed quartic form without the $y^3$ term. 3. **Step 1: Substitution** Substitute $x = y - \frac{b}{4}$ into the original equation: $$\left(y - \frac{b}{4}\right)^4 + b \left(y - \frac{b}{4}\right)^3 + c \left(y - \frac{b}{4}\right)^2 + d \left(y - \frac{b}{4}\right) + e = 0$$ 4. **Step 2: Expand powers** Expand each term using binomial expansion: $$\left(y - \frac{b}{4}\right)^4 = y^4 - b y^3 + \frac{3b^2}{8} y^2 - \frac{b^3}{16} y + \frac{b^4}{256}$$ $$\left(y - \frac{b}{4}\right)^3 = y^3 - \frac{3b}{4} y^2 + \frac{3b^2}{16} y - \frac{b^3}{64}$$ $$\left(y - \frac{b}{4}\right)^2 = y^2 - \frac{b}{2} y + \frac{b^2}{16}$$ 5. **Step 3: Substitute expansions back** Plug these into the equation: $$y^4 - b y^3 + \frac{3b^2}{8} y^2 - \frac{b^3}{16} y + \frac{b^4}{256} + b \left(y^3 - \frac{3b}{4} y^2 + \frac{3b^2}{16} y - \frac{b^3}{64}\right) + c \left(y^2 - \frac{b}{2} y + \frac{b^2}{16}\right) + d \left(y - \frac{b}{4}\right) + e = 0$$ 6. **Step 4: Combine like terms** Group terms by powers of $y$: - $y^4$ term: $y^4$ - $y^3$ terms: $-b y^3 + b y^3 = 0$ (cubic terms cancel) - $y^2$ terms: $\frac{3b^2}{8} y^2 - \frac{3b^2}{4} y^2 + c y^2 = \left(c - \frac{3b^2}{8}\right) y^2$ - $y$ terms: $- \frac{b^3}{16} y + \frac{3b^3}{16} y - \frac{b c}{2} y + d y = \left(d - \frac{b c}{2} + \frac{b^3}{8}\right) y$ - Constant terms: $\frac{b^4}{256} - \frac{b^4}{64} + \frac{c b^2}{16} - \frac{b d}{4} + e = e - \frac{b d}{4} + \frac{c b^2}{16} - \frac{3 b^4}{256}$ 7. **Step 5: Final depressed quartic form** The equation becomes: $$y^4 + \left(c - \frac{3b^2}{8}\right) y^2 + \left(d - \frac{b c}{2} + \frac{b^3}{8}\right) y + \left(e - \frac{b d}{4} + \frac{c b^2}{16} - \frac{3 b^4}{256}\right) = 0$$ This is a quartic without the cubic term, which is easier to solve. **Summary:** The substitution $x = y - \frac{b}{4}$ removes the cubic term from the quartic equation, yielding a depressed quartic: $$y^4 + p y^2 + q y + r = 0$$ where $$p = c - \frac{3b^2}{8}, \quad q = d - \frac{b c}{2} + \frac{b^3}{8}, \quad r = e - \frac{b d}{4} + \frac{c b^2}{16} - \frac{3 b^4}{256}$$ This form is the basis for solving quartic equations using Ferrari's method or other techniques.