How do you factor #x^2+11=300#?

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Answer 1 · 2015-09-16T15:55:09

#(x+17)(x-17)=0#

Subtract #300# from both sides

#x^2+11-300=300-300#

#x^2-289=0#

The left hand side is the difference of two squares.
We factor as follows

#(x+17)(x-17)=0#

Or we could say

#x^2=289# by subtracting #11# from both sides

Taking the square root we have

#x=+-sqrt(289)=+-17#

but you want the factorization so we go with what i did first

Answer 2 · 2015-09-16T15:55:15

(x-17)(x+17)

To factorise this, recollect that #17^2=289#

The give expression can be simplified to #x^2 -300 +11=0#

#x^2-289=0#
#(x^2- 17^2)=0#

(x-17)(x+17)=0