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

Dec 24, 2016

$16 {x}^{2} + 8 x + 1 = {\left(4 x + 1\right)}^{2}$

#### Explanation:

Notice that:

$1681 = {41}^{2}$

Hence we find:

$16 {x}^{2} + 8 x + 1 = {\left(4 x + 1\right)}^{2}$

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

$16 {x}^{2} + 8 x + 1 = 16 {\left(10\right)}^{2} + 8 \left(10\right) + 1 = 1600 + 80 + 1 = 1681$

$4 x + 1 = 4 \left(10\right) + 1 = 40 + 1 = 41$

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 $16 {x}^{2} = {\left(4 x\right)}^{2}$ and $1 = {1}^{2}$ are perfect squares. So does the middle term match when we square $\left(4 x + 1\right)$ ?

${\left(4 x + 1\right)}^{2} = {\left(4 x\right)}^{2} + 2 \left(4 x\right) \left(1\right) + {1}^{2} = 16 {x}^{2} + 8 x + 1 \text{ }$ - Yes.

In general:

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

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