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.

Related questions
See all questions in Factorization of Quadratic Expressions
Impact of this question
1413 views around the world
You can reuse this answer
Creative Commons License