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.

How do you factor 16x^2+8x+116x^2+8x+1?