How do you factor the trinomial # 49x^2 − 8x + 16#?

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Answer 1 · 2015-12-22T22:59:11

You use the formula : #(a-b)^2 = a^2-2ab+b^2#

#a = 7x#
#b = -4/7#

#a^2=49x^2#
#b^2= 16/49#
#2ab = -8x#

#49x^2-8x+16/49-16/49+16#

#(7x-4/7)^2+768/49#

Answer 2 · 2015-12-23T00:03:25

#49x^2-8x+16#

#= (x-4/49-(16sqrt(3)i)/49)(x-4/49+(16sqrt(3)i)/49)#

#49x^2-8x+16# is of the form #ax^2+bx+c# with #a=49#, #b=-8# and #c=16#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-8)^2-(4xx49xx16) = 64-3136 = -3072#

#= -3*2^10#

Since this is negative, our trinomial has no linear factors with Real coefficients.

It has Complex zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (8+-sqrt(Delta))/98#

#=(8+-32sqrt(3)i)/98 = (4+-16sqrt(3)i)/49#

Hence:

#49x^2-8x+16#

#= (x-4/49-(16sqrt(3)i)/49)(x-4/49+(16sqrt(3)i)/49)#