How do you subtract #3x^2-4# from the sum of #x^2-6x+1# and #2x^2-5x+5#?

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Answer 1 · 2016-11-09T22:41:28

#-11x + 10#

The sum of #x^2 - 6x + 1# and #2x^2 - 5x + 5# is solved as:

#x^2 - 6x + 1 + 2x^2 - 5x + 5 =>#

#x^2 + 2x^2 - 6x - 5x + 1 + 5 =>#

#3x^2 - 11x + 6 =>#

Then to subtract #3x^2 - 4# from this is solved as:

#3x^2 - 11x + 6 - (3x^2 - 4) =>#

#3x^2 - 11x + 6 - 3x^2 + 4) =>#

#3x^2 - 3x^2 -11x + 6 + 4#

#-11x + 10#

Answer 2 · 2016-11-09T22:50:15

#" "-11x+10#

Using the old fashioned approach of lining things up.

The sum of #(x^2-6x+1) +(2x^2-5x+5)#

#color(white)(2)x^2-6x+1#
#ul(2x^2-5x+5)#
#color(red)(3x^2-11x+6)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now subtract #color(green)(3x^3-4)#

I am using the place holder of no values of #x# as #0x# to make things line up properly.

#color(red)(3x^2-11x+6)#
#color(green)(ul(3x^2+color(white)(1)0x-4)) larr" Subtract"#
#0x^2-11x+10#

Giving: #" "-11x+10#