Can someone explain why, when factored, both terms are x−x- (circled in red). I got the x term just fine, but am confused as to why both are minus signs after the x.
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It is a theorem about polynomials that x minus any root of f(x) is ALWAYS a factor of f(x) if f(x) is a polynomial.
[MATH]f(x) =ax^2 + bx + c, \ u = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}, \text { and } v = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}.[/MATH]
[MATH]\therefore f(u) = a \left ( \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \right )^2 + b \left ( \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \right ) + c =[/MATH]
[MATH]a \left ( \dfrac{b^2 - 2b\sqrt{b^2 - 4ac} + b^2 - 4ac}{4a^2} \right ) + \dfrac{-b^2 + b\sqrt{b^2 - 4ac}}{2a} + c =[/MATH]
[MATH]\dfrac{2b^2 - 2b\sqrt{b^2 - 4ac} - 4ac}{4a} + \dfrac{-2b^2 + 2b\sqrt{b^2 - 4ac}}{4a} + \dfrac{4ac}{4a} =[/MATH]
[MATH]\dfrac{2b^2 - 2b^2 + 2b\sqrt{b^2 - 4ac} - 2b\sqrt{b^2 - 4ac} +4ac - 4ac}{4a} = \dfrac{0 + 0 + 0}{4} = 0.[/MATH]
I shall let you satisfy yourself that f(v) = 0.
[MATH]a \left ( x - \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \right ) \left ( x - \dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \right ) =[/MATH]
[MATH]ax^2 - \dfrac{-bx + x\sqrt{b^2 - 4ac}}{2} - \dfrac{-bx - x\sqrt{b^2 - 4ac}}{2} + \dfrac{b^2 - b^2 + 4ac}{4a} =[/MATH]
[MATH]ax^2 - \dfrac{2bx}{2} + \dfrac{4ac}{4a} = ax^2 + bx + c.[/MATH]
It is a specific application of the Fundamental Theorem of Algebra to quadratics.