How do you factor #16x^2+8x+1#?
1 Answer
Dec 24, 2016
Explanation:
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
#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
Another way we could spot this is as follows:
Notice that both
#(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