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

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

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Answer 1 · 2016-06-16T20:49:09

${(4 x - 1)}^{2}$ $(4x-1)^2$

Explanation:

This is a polynomial of second degree and since the coefficient of
$" "$
$x < 0$ $x<0$ , we think of the binomial property that says:
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$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$
$" "$
In the given polynomial first term $16 x^{2} = {(4 x)}^{2}$ $16x^2=(4x)^2$ and $1 = {(1)}^{2}$ $1=(1)^2$
$" "$
$16 x^{2} - 8 x + 1$ $16x^2-8x+1$
$" "$
$= {(4 x)}^{2} - 2 (4 x) (1) + 1^{2}$ $=(4x)^2-2(4x)(1)+1^2$
$" "$
$= {(4 x - 1)}^{2}$ $=(4x-1)^2$

Answer 2 · 2017-09-29T11:15:53

$16 x^{2} - 8 x + 1$ $16x^2 -8x+1$

$= (4 x - 1) (4 x - 1)$ $=(4x-1)(4x-1)$

Explanation:

The $+$ $color(lime)(+)$ sign in the third term indicates two things:

the factors need to be $A D D E D$ $color(lime)(ADDED)$
the signs in the brackets will $be the same$ $color(lime)("be the same")$

The $-$ $color(red)(-)$ sign in the second term indicates that the signs will be negative.

$16 x^{2} - 8 x + 1$ $16x^2color(red)(-)8x color(lime)(+) 1$

Find factors of $16 and 1$ $16 and 1$ which add to $8$ $8$ .

The factors of $1$ $1$ are just $1$ $1$ , so we can ignore them.

The factors of $16$ $16$ which add to $8$ $8$ are $4 and 4$ $4 and 4$

$4 \times 4 = 16 and 4 + 4 = 8$ $4xx4 = 16 and 4+4=8$

$16 x^{2} - 8 x + 1$ $16x^2 -8x+1$

$= (4 x - 1) (4 x - 1)$ $=(4x-1)(4x-1)$

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

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