How do you factor #x^2 - 8x + 16#?

Question

49082 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-06T14:09:01

#color(green)(=> (x-4)^2#

#x^2 - 8x + 16#

#=>x^2 - (2 * 4 * x ) + 4^2#

It’s in the form #a^2 - 2 ab + b^2 = (a- b)^2#

#:. => (x-4)^2#

Answer 2 · 2018-07-06T14:09:12

#(x-4)^2#

Given: #x^2-8x+16#.

We find two numbers, say #a# and #b#, such that #a+b=-8# and #ab=16#.

Mentally, I see that #a=b=-4#. Checking, #(-4)+(-4)=-8,-4*-4=16#.

Now, we split it into:

#=x^2+ax+bx+16#

#=x^2-4x-4x+16#

#=x(x-4)-4(x-4)#

#=(x-4)(x-4)#

#=(x-4)^2#