Transcription of Solving Cubic Polynomials - SHSU
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Solving Cubic Polynomials - SHSU
Solving Cubic The general solution to the quadratic equationThere are four steps to finding the zeroes of a quadratic First divide by the leading term, making the Then, givenx2+a1x+a0, substitutex=y a12to obtain an equation without the linear term.(This is the depressed equation.)3. Solve then foryas a square root. (Remember to use both signs of the square root.)4. Once this is done, recoverxusing the fact thatx=y example, let s solve2x2+ 7x 15 = , we divide both sides by 2 to create an equation with leading term equal to one:x2+72x 152= replacexbyx=y a12=y 74to obtain:y2=16916 Solve fory:y=134or 134 Then, Solving back forx, we havex=32or method is equivalent to completing the square and is the steps taken in developing the much-memorized quadratic formula.
This does not change the roots. 2.Then, given xn+a n 1x n1 +a n 2x 2 +:::a 1x+a 0, substitute x= y a n 1 n to obtain an equation without the term of degree n 1:(This is the depressed polynomial.) Since this step is reversible, solutions to the \depressed equation" give us solutions to …
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