How do you write # y=1.2(x+10)^2+6# into factored form?

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Answer 1 · 2015-06-02T20:37:01

#(x+10)^2 >= 0#, so #y >= 6# for all #x in RR#.

Since there are no real values of #x# for which #y = 0#,
#y = 6/5((x+10)^2-5)# has no linear factors with real coefficients.

Is there an error in the question?

How about #y = 1.2(x+10)^2 - 6# ?

Then #y = 6/5((x+10)^2-5)# when #x+10 = +-sqrt(5)#

That is #y = 0# when #x = -10+-sqrt(5)#

Hence

#y = 1.2(x+10)^2-6#

#= 6/5(x+10-sqrt(5))(x+10+sqrt(5))#