How do you factor given that #f(6)=0# and #f(x)=x^3-3x^2-16x-12#?

1 Answer
Jim G.
Sep 4, 2017

#f(x)=(x-6)(x+1)(x+2)#

Explanation:

#"using the "color(blue)"factor theorem"#

#•color(white)(x)f(a)=0" if and only if "(x-a)" is a factor of "f(x)#

#f(6)=0rArr(x-6)" is a factor of "f(x)#

#f(x)=color(red)(x^2)(x-6)color(magenta)(+6x^2)-3x^2-16x-12#

#color(white)(f(x))=color(red)(x^2)(x-6)color(red)(+3x)(x-6)color(magenta)(+18x)-16x-12#

#color(white)(f(x))=color(red)(x^2)(x-6)color(red)(+3x)(x-6)color(red)(+2)(x-6)color(magenta)(+12)-12#

#color(white)(f(x))=color(red)(x^2)(x-6)color(red)(+3x)(x-6)color(red)(+2)(x-6)+0#

#rArrf(x)=(x-6)(color(red)(x^2+3x+2))#

#color(white)(rArrf(x))=(x-6)(x+1)(x+2)#

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