How do you factor the trinomial #49x^2 + 56x +9#?

1 Answer
Alan P.
Feb 1, 2016

Nothing pretty; I got
#color(white)("XXX")(x+(28+sqrt(343))/49)(x+(28-sqrt(343))/49)#

Explanation:

You might check the question; if the last term were #(-9)# instead of #(+9)# then there would be some nice integer terms.

Working with what is given, my only approach would be to apply the quadratic formula:
#color(white)("XXX")x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)#
to the general equation form: #ax^2+bx+c#
to get the factors #(x-x_1)(x-x_2)#

In this case
#color(white)("XXX")x_(1,2) = (-56+-sqrt(56^2-4*49*9))/(2*49)#

#color(white)("XXXXXX")=(-56+-sqrt(3136-1764))/(2*49)#

#color(white)("XXXXXX")=(-56+-sqrt(1372))/(2*49)#

#color(white)("XXXXXX")=(-56+-2sqrt(343))/(2*49)#

#color(white)("XXXXXX")=(-28+-sqrt(343))/49#

So the factors are
#color(white)("XXX")(x-((-28-sqrt(343))/49))*(x-((-28+sqrt(343))/49))#

...all this assumes that I didn't make any arithmetic errors.

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