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.

Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to nding the zeroes of a quadratic polynomial. 1.First divide by the leading term, making the polynomial monic. 2.Then, given x2 + a 1x+ a 0, substitute x= y a 1 2 to obtain an equation without the linear term. (This is the \depressed" equation.)

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