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. For example, if the original equation is our high school quadratic ax2+bx+c= 0then the first step creates the equationx2+bax+ca= then writex=y b2aand obtain, after simplifying,y2 b2 4ac4a2= 0so thaty= b2 4ac2aand sox= b2a b2 solutions to this quadratic depend heavily on the value ofb2 4ac.
We teach a version of this method in high school when students learn to solve quadratic equations by factoring. For example, one might solve the equation 3x2 2x 8 = 0 by factoring the left-hand side into (3x+ 4)(x 2), obtain solutions x= 4 …
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