{"id":203,"date":"2025-12-02T09:12:47","date_gmt":"2025-12-02T09:12:47","guid":{"rendered":"https:\/\/potentsky.com\/math\/?page_id=203"},"modified":"2025-12-02T10:57:52","modified_gmt":"2025-12-02T10:57:52","slug":"203-2","status":"publish","type":"page","link":"https:\/\/potentsky.com\/math\/203-2\/","title":{"rendered":"Depressed Polynomials"},"content":{"rendered":"\ncompleting the square method gives us a quadratic formua, why can’t completing a cube give us a cubic formulaWhile a form of “completing the cube” is the essential first step in solving a cubic equation, it cannot, by itself, produce a general formula in the same direct way as completing the square does for quadratics. The resulting equation is not as simple as merely taking a cube root.\u00a0The Limitation of Completing the Cube\u00a0“Completing the square” works because it transforms the general quadratic equation \\(ax^{2}+bx+c=0\\) into the simple form \\((x+s)^{2}=t\\) (a perfect square equal to a constant), which is trivial to solve by taking the square root.\u00a0For a cubic equation \\(ax^{3}+bx^{2}+cx+d=0\\), the initial step in the general solution method (known as Cardano’s method) is to “complete the cube” in a specific way using a substitution \\(x=y-\\frac{b}{3a}\\). This substitution successfully eliminates the \\(y^{2}\\) (quadratic) term, resulting in a depressed cubic of the form \\(y^{3}+py+q=0\\).\u00a0However, unlike the quadratic case where all variable terms are absorbed into the perfect square, the depressed cubic still has two variable terms (\\(y^{3}\\) and \\(py\\)). The equation cannot be reduced to a form like \\((y+s)^{3}=t\\) for arbitrary coefficients \\(p\\) and \\(q\\). A general cubic equation has three independent coefficients (after normalization), while a simple perfect cube equation \\((y+s)^{3}=t\\) only has two free variables (\\(s\\) and \\(t\\)).\u00a0How the Cubic Formula Is Actually Derived\u00a0The full cubic formula requires further, more complex substitutions to solve the depressed cubic. It involves:\u00a0Transforming the general cubic into the depressed form \\(y^{3}+py+q=0\\).Using a secondary substitution (Vieta’s substitution, \\(y=u+v\\)) to transform the equation into a quadratic equation in terms of \\(u^{3}\\).Solving this new quadratic using the familiar quadratic formula to find values for \\(u^{3}\\) (and consequently \\(v^{3}\\)).Taking the cube roots of \\(u^{3}\\) and \\(v^{3}\\) (remembering there are three complex cube roots for each) to find \\(u\\) and \\(v\\).Combining these to get the solutions for \\(y\\), and then back-substituting to find the solutions for \\(x\\).\u00a0The resulting formula is much longer, more complicated, and generally not taught in schools due to its complexity and the fact that it often requires working with complex numbers even when the final roots are all real (a situation called casus irreducibilis).\n\n\n\nCan you apply the same principles of depressing the cubic to depress a quadratic equationYes, you can apply the same principle used to depress a cubic equation to “depress” a quadratic equation.\u00a0Depressing a polynomial means using a change of variables (a substitution) to eliminate the second-highest degree term. For a general quadratic equation of the form:\u00a0\\(ax^{2}+bx+c=0\\)You can eliminate the \\(bx\\) term by using a substitution that shifts the variable \\(x\\) by a specific amount.\u00a0The Process\u00a0Normalize the equation: Divide all terms by \\(a\\) to get a leading coefficient of 1:\\(x^{2}+Px+Q=0\\)where \\(P=b\/a\\) and \\(Q=c\/a\\).Apply the substitution: Just as the cubic uses \\(x=y-\\frac{P}{3}\\), the quadratic uses the substitution:\\(x=y-\\frac{P}{2}\\)The value \\(\\frac{P}{2}\\) corresponds to \\(\\frac{b}{2a}\\) from the standard quadratic formula derivation.Substitute and simplify: Substitute \\(x=y-\\frac{P}{2}\\) into the normalized equation:\\(\\left(y-\\frac{P}{2}\\right)^{2}+P\\left(y-\\frac{P}{2}\\right)+Q=0\\)Expand the terms:\\(\\left(y^{2}-Py+\\frac{P^{2}}{4}\\right)+\\left(Py-\\frac{P^{2}}{2}\\right)+Q=0\\)Combine like terms:\\(y^{2}+(-P+P)y+\\left(\\frac{P^{2}}{4}-\\frac{P^{2}}{2}+Q\\right)=0\\)\\(y^{2}+0y+\\left(-\\frac{P^{2}}{4}+Q\\right)=0\\)\u00a0The Result: The Depressed Quadratic\u00a0The result is the depressed quadratic equation:\u00a0\\(y^{2}+\\left(Q-\\frac{P^{2}}{4}\\right)=0\\)or simply:\u00a0\\(y^{2}=\\frac{P^{2}}{4}-Q\\)This depressed form is precisely the “completing the square” result, allowing you to easily solve for \\(y\\) by taking the square root of both sides.\u00a0\\(y=\\pm \\sqrt{\\frac{P^{2}}{4}-Q}\\)Then you back-substitute using \\(x=y-\\frac{P}{2}\\) to find the final solutions for \\(x\\), deriving the standard quadratic formula.\u00a0So, while the term “depressing the quadratic” isn’t commonly used in introductory algebra (we call it “completing the square”), it utilizes the exact same principle as depressing the cubic to eliminate the second-degree term.\n\n\n\nCan it be used quartic general formulaYes, the same principle of depressing the polynomial is the essential first step in finding the general quartic (degree 4) formula. This method, developed by the Italian mathematician Lodovico Ferrari in the 16th century, is the standard algebraic approach to solving quartic equations.\u00a0The Process for a Quartic Equation\u00a0For a general quartic equation of the form:\u00a0\\(Ax^{4}+Bx^{3}+Cx^{2}+Dx+E=0\\)\u00a0Normalize: Divide the entire equation by \\(A\\) to get a leading coefficient of 1.Depress the Equation: Eliminate the \\(x^{3}\\) term by using the substitution \\(x=y-\\frac{B}{4A}\\). This results in a depressed quartic equation that only has terms in \\(y^{4}\\), \\(y^{2}\\), \\(y\\), and a constant (no \\(y^{3}\\) term).Introduce an Auxiliary Variable (Ferrari’s Method): The next steps are more complex than the cubic method. You rearrange the depressed quartic to group terms and try to form a perfect square on one side. This introduces a new, auxiliary variable (often \\(\\lambda \\) or \\(y\\)).Solve a Resolvent Cubic: The process requires choosing a specific value for this new variable that makes the other side of the equation also a perfect square. The condition to make this second side a perfect square is a cubic equation in terms of the auxiliary variable.Solve Two Quadratics: You solve this “resolvent cubic” using the cubic formula. Once you have a solution from the cubic equation, you can factor the depressed quartic into two quadratic equations and solve each one using the quadratic formula.Back-substitute: Finally, reverse the initial substitution \\(x=y-\\frac{B}{4A}\\) to find the four roots of the original general quartic equation.\u00a0The full quartic formula, when written out explicitly, is extremely long and impractical to use by hand due to its complexity and the nesting of square and cube roots. The key takeaway is that the solution relies on reducing the problem to the solution of lower-degree polynomials (a cubic, then two quadratics), starting with the depression step.\n\n\n\nGetting the leading coefficient to be 1 is essential in depressing the polynomials; can it be avoidedWhile normalizing the leading coefficient to 1 (\\(A=1\\)) makes the algebraic steps for depressing a polynomial cleaner and conceptually simpler to derive the general formulas, it is not strictly essential or unavoidable.\u00a0You can perform the substitution using the original coefficients directly, but the resulting algebra becomes more cumbersome.\u00a0Why the \\(A=1\\) Normalization is Standard\u00a0The primary reason for dividing by the leading coefficient \\(A\\) is simplification:\u00a0Simpler Substitution Term: For a general polynomial \\(Ax^{n}+Bx^{n-1}+\\dots =0\\), the necessary substitution to eliminate the \\(x^{n-1}\\) term is always \\(x=y-\\frac{B}{nA}\\). Dividing by \\(A\\) first simplifies this fraction to \\(x=y-\\frac{P}{n}\\) (where \\(P=B\/A\\)).Cleaner Derived Coefficients: Working with integer or simpler fractional coefficients usually leads to fewer calculation errors during the expansion phase.Standardized Formulas: The general formulas for the quadratic, cubic, and quartic are typically published using the normalized coefficients \\(P,Q,R,\\dots \\), making \\(A=1\\) a standard convention.\u00a0How to Avoid Normalization\u00a0You could theoretically work with the original \\(A\\) present throughout the entire process. The substitution for the general quadratic \\(Ax^{2}+Bx+C=0\\) would still be \\(x=y-\\frac{B}{2A}\\).\u00a0The initial expansion would look more complicated, as the \\(A\\) factor would multiply every term:\u00a0\\(A\\left(y-\\frac{B}{2A}\\right)^{2}+B\\left(y-\\frac{B}{2A}\\right)+C=0\\)After careful expansion and simplification, the \\(y^{1}\\) term would still successfully cancel out, and you would arrive at the same depressed form, just with \\(A\\) multiplying the entire equation:\u00a0\\(A\\cdot y^{2}+\\left(\\text{constant\\ terms}\\right)=0\\)Dividing by \\(A\\) at this later stage gives the exact same result as dividing at the beginning.\u00a0In conclusion, while you can technically avoid the division until the end, normalizing the equation at the start is a procedural shortcut that makes the intermediate derivation steps manageable.\n\n\n\nCan authoritatively conclude that the concept of dividing by A is to make the first term of the polynomial a perfect squareIt is more accurate to conclude that dividing by \\(A\\) makes the leading coefficient equal to 1, which is a necessary step to eliminate the second-highest degree term through the variable substitution (the “depressing” step).\u00a0The division is done to make the resulting algebra of the depressing substitution simple enough to manage, not specifically to make the first term a perfect square.\u00a0Here’s why:\u00a0For the quadratic (\\(ax^{2}+bx+c=0\\)): Dividing by \\(a\\) results in \\(x^{2}+Px+Q=0\\). The process of “completing the square” works after this step, transforming the first two terms into a perfect square plus a constant: \\(x^{2}+Px=(x+P\/2)^{2}-(P\/2)^{2}\\). The division by \\(a\\) sets up the ability to easily complete the square.For higher-degree polynomials (cubic, quartic, etc.): The division by \\(A\\) allows the use of the standard, simple substitution \\(x=y-B\/(nA)\\) which guarantees the elimination of the \\(y^{n-1}\\) term. It is a normalization step, not a “perfect square” step. The goal is term elimination, which subsequently allows other, more complex algebraic maneuvers to find the roots.\n\n\n\nSo it essential to conclude that we divide polynomials by A to make the first term perfect polynomial as in square, cube, quartic etcNo, that conclusion is incorrect. The division by the leading coefficient \\(A\\) is done to normalize the polynomial so that the coefficient of the highest-degree term is 1, not to make the first term a “perfect polynomial” or a perfect square, cube, or quartic.\u00a0The division by \\(A\\) serves two distinct purposes in the context of solving polynomial equations:\u00a0Normalization for Term Elimination: The primary goal in solving cubic and quartic equations is to eliminate the second-highest degree term (the \\(x^{n-1}\\) term) using the substitution \\(x=y-B\/(nA)\\). This substitution only works neatly and consistently after you have divided by \\(A\\).Setting up “Completing the Square”: In the specific case of the quadratic equation, dividing by \\(A\\) yields \\(x^{2}+Px+Q=0\\). This specific format is necessary to apply the “completing the square” algebraic manipulation, which forms a perfect square binomial, \\((x+P\/2)^{2}\\).\u00a0For cubics and quartics, you do not immediately form a perfect cube or quartic after dividing by \\(A\\). You only eliminate the second term, which is the necessary step to apply the more complex methods (like Ferrari’s or Cardano’s methods) that follow.\n\n\n\nBut still my observation seems correct though not the main objective of dividing by A throughout the polynomial equationYour observation is correct as a consequence of the division: the resulting leading term is \\(x^{n}\\), which has a coefficient of 1.\u00a0Mathematically, a polynomial with a leading coefficient of 1 is called a monic polynomial.\u00a0So, the result of dividing the polynomial by \\(A\\) is indeed a monic polynomial where the leading term is a “perfect” power of the variable (\\(x^{2},x^{3},x^{4}\\), etc.).\u00a0However, the objective of this step in the context of solving general formulas (quadratic, cubic, quartic) remains the simplification of the subsequent algebraic steps required to eliminate the second-highest degree term.\n\n\n\nso with quadratic equations perfection is achieved more immediately than higher degree polynomialsYes, that is a precise way to summarize the distinction.\u00a0With quadratic equations, once you normalize the equation by dividing by \\(A\\) and rearrange the terms, the “perfection” of a perfect square binomial is achieved immediately and directly yields the solution upon taking a square root. The depressed form \\(y^{2}=\\text{constant}\\) is instantly solvable.\u00a0For higher-degree polynomials (cubic, quartic):\u00a0The initial normalization and depression step does create a monic polynomial with the second term eliminated.However, this depressed form (\\(y^{3}+py+q=0\\) or \\(y^{4}+py^{2}+qy+r=0\\)) is not immediately a “perfect” power of a binomial. It still contains multiple variable terms.Achieving the final solution requires further, more complex algebraic “tricks” (like introducing a second variable or solving an intermediate “resolvent” cubic equation) before you can isolate the variable using roots.\u00a0The simplicity of the quadratic formula lies in this immediate achievement of a perfect square structure.\n\n\n\nThe same way derivatives are applied in cubic equation, can it be applied in quadratic equation even if not necessary and arrive at te same answerThinkingYes, derivatives can absolutely be applied to a quadratic equation to find a key piece of information that helps solve the equation: the vertex. While it is an unnecessary step compared to simply completing the square, using the derivative will lead you to the same solution.\u00a0Applying Derivatives to a Quadratic\u00a0The derivative of a function tells you the slope of the tangent line at any point. For a smooth curve like a parabola (a quadratic equation’s graph), the vertex (either the maximum or minimum point) occurs exactly where the slope is zero.\u00a0Here is how you apply it to the general quadratic equation \\(f(x)=ax^{2}+bx+c\\):\u00a0Find the first derivative of the function:Using basic differentiation rules (the power rule):\\(f^{\\prime }(x)=\\frac{d}{dx}(ax^{2}+bx+c)\\)\\(f^{\\prime }(x)=2ax+b\\)Set the derivative to zero to find the critical point (the \\(x\\)-coordinate of the vertex):\\(2ax+b=0\\)\\(2ax=-b\\)\\(x=-\\frac{b}{2a}\\)\u00a0Arriving at the Same Answer (Vertex and the Formula)\u00a0This result, \\(x=-\\frac{b}{2a}\\), is the exact \\(x\\)-coordinate of the vertex and the same value derived as the “center” of the solution in the completing the square method. It is the symmetry point of the parabola.\u00a0In the context of the general quadratic formula derivation:\u00a0Completing the Square Method: You solve for \\(x\\) directly by isolating the variable terms, leading to the full formula: \\(x=\\frac{-b\\pm \\sqrt{b^{2}-4ac}}{2a}\\).Derivative Method: The value \\(x=-\\frac{b}{2a}\\) is the midpoint between the two roots. To get the actual roots (where \\(f(x)=0\\)), you would still need to plug this \\(x\\)-value back into the original equation to find the \\(y\\)-value of the vertex, and then use the resulting depressed form to find the remaining \\(\\pm \\) component (the \\(\\sqrt{b^{2}-4ac}\\) part).\u00a0So the derivative method helps you find the vertex quickly, which is a key part of the solution structure, but you still need the algebraic steps of the original formula derivation to find the final two roots.\u00a0 \n","protected":false},"excerpt":{"rendered":"

completing the square method gives us a quadratic formua, why can’t completing a cube give us a cubic formulaWhile a form of “completing the cube” is the essential first step in solving a cubic equation, it cannot, by itself, produce a general formula in the same direct way as completing the square does for quadratics. 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