How do you factor #16x^2+8x+1#?

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Answer 1 · 2016-12-24T12:21:14

#16x^2+8x+1 = (4x+1)^2#

Notice that:

#1681 = 41^2#

Hence we find:

#16x^2+8x+1 = (4x+1)^2#

Was that a bit fast? Think what happens when we put #x=10#:

#16x^2+8x+1 = 16(10)^2+8(10)+1 = 1600+80+1 = 1681#

#4x+1 = 4(10)+1 = 40+1 = 41#

When we square #41# the only carry is in the most significant digits, so this 'trick' works for this example.

Another way we could spot this is as follows:

Notice that both #16x^2 = (4x)^2# and #1 = 1^2# are perfect squares. So does the middle term match when we square #(4x+1)# ?

#(4x+1)^2 = (4x)^2+2(4x)(1)+1^2 = 16x^2+8x+1" "# - Yes.

In general:

#(a+b)^2 = a^2+2ab+b^2#

So if we can identify #a# and #b# then we just require the middle term to be twice the product.