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A weird method for factoring quadratics (and why it works) 1 год назад


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A weird method for factoring quadratics (and why it works)

Figuring out the inner workings of a strange but useful factoring algorithm I learned as a kid. End of the video - how to prove that the scaling relationship is true: There are two ways I can think of to go about this. First, if you just care about showing that the roots of the new equation are "a" times as large, write out the quadratic formula to find the zeros of each and compare the results. If you want to show that the entire new parabola (not just the roots) is a scaled version of the original, we can scale the original by a factor of "a" along the x-axis by taking y = ax^2 + bx + c and substituting (x/a) for all instances of x. Likewise, to scale it along the y-axis, substitute (y/a) for y. You can show that this process gets you to the equation y = x^2 + bx + ac after simplifying.

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