How do you factor $16 x^{2} + 8 x + 1$ $16x^2+8x+1$ ?

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How do you factor $16 x^{2} + 8 x + 1$ $16x^2+8x+1$ ?

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Answer 1 · 2016-12-24T12:21:14

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

Explanation:

Notice that:

$1681 = 41^{2}$ $1681 = 41^2$

Hence we find:

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

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

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

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

When we square $41$ $41$ 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 $16 x^{2} = {(4 x)}^{2}$ $16x^2 = (4x)^2$ and $1 = 1^{2}$ $1 = 1^2$ are perfect squares. So does the middle term match when we square $(4 x + 1)$ $(4x+1)$ ?

${(4 x + 1)}^{2} = {(4 x)}^{2} + 2 (4 x) (1) + 1^{2} = 16 x^{2} + 8 x + 1$ $(4x+1)^2 = (4x)^2+2(4x)(1)+1^2 = 16x^2+8x+1" "$ - Yes.

In general:

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2 = a^2+2ab+b^2$

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

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How do you factor $16 x^{2} + 8 x + 1$ $16x^2+8x+1$ ?

1 Answer

Explanation:

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