How do you factor completely #4r^2 -21+5#?

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Answer 1 · 2016-06-20T15:55:09

#(2r+4)(2r-4)#

To factor #4r^2-21+5# completely we would begin by combining like terms.

#4r^2-16#

We next would recognize that the binomial is a difference (subtraction) of two squares.

#4r^2 = (2r)(2r)#
and
#16 = (4)(4)#

The difference of two squares always factors in this pattern

#a^2-b^2#

#(a+b)((a-b)#

So our binomial doctors as

#4r^2-16#

#(2r+4)(2r-4)#