Question 193037
a) The given statement is true, here's why...



*[Tex \LARGE a(x-\alpha)(x-\beta)] ... Start with the given expression.



*[Tex \LARGE a(x^2-\alpha x-\beta x +\alpha\beta)] ... FOIL



*[Tex \LARGE a(x^2-(\alpha+\beta) x +\alpha\beta)] ... Combine like terms.



Let *[Tex \LARGE b=-(\alpha+\beta)] and *[Tex \LARGE c=\alpha\beta] to get



*[Tex \LARGE a(x^2+bx+c)]



*[Tex \LARGE a\left(x^2+bx+\frac{b^2}{4}-\frac{b^2}{4}+c\right)] ... Take half of "b" and square it to get {{{b^2/4}}}. Add AND subtract this inside the parenthesis.



*[Tex \LARGE a\left(\left(x+\frac{b}{2}\right)^2-\frac{b^2}{4}+c\right)] ... Factor the first three terms in the parenthesis



*[Tex \LARGE a\left(x+\frac{b}{2}\right)^2+a\left(-\frac{b^2}{4}+c\right)] ... Distribute



Let *[Tex \LARGE k=a\left(-\frac{b^2}{4}+c\right)] to get



*[Tex \LARGE a\left(x+\frac{b}{2}\right)^2+k]



Let *[Tex \LARGE h=-\frac{b}{2}] to get 



*[Tex \LARGE a\left(x-h\right)^2+k]



So for ANY expression of the form *[Tex \LARGE a(x-\alpha)(x-\beta)] you can rewrite it as *[Tex \LARGE a\left(x-h\right)^2+k]




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b) The given statement is false (if you are only restricted to factor over the reals)



Here's a counter-example:



Let {{{a=1}}}, {{{h=1}}}, and {{{k=3}}}. So the general expression {{{a(x-h)^2+k}}} becomes



{{{(x-1)^2+3}}}


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{{{(x-1)^2+3}}} Start with the given expression.



{{{x^2-2x+1+3}}} FOIL



{{{x^2-2x+4}}} Combine like terms.



Since you CANNOT factor {{{x^2-2x+4}}} over the reals, this means that {{{(x-1)^2+3}}} CANNOT be written in the form of *[Tex \LARGE a(x-\alpha)(x-\beta)] 



Note: if you are allowed to factor over the complex numbers, then you can rewrite {{{a(x-h)^2+k}}} into *[Tex \LARGE a(x-\alpha)(x-\beta)]