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

Jun 16, 2016

${\left(4 x - 1\right)}^{2}$

#### Explanation:

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

Sep 29, 2017

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

$= \left(4 x - 1\right) \left(4 x - 1\right)$

#### Explanation:

The $\textcolor{\lim e}{+}$ sign in the third term indicates two things:

• the factors need to be $\textcolor{\lim e}{A D D E D}$
• the signs in the brackets will $\textcolor{\lim e}{\text{be the same}}$

The $\textcolor{red}{-}$sign in the second term indicates that the signs will be negative.

$16 {x}^{2} \textcolor{red}{-} 8 x \textcolor{\lim e}{+} 1$

Find factors of $16 \mathmr{and} 1$ which add to $8$.

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

The factors of $16$ which add to $8$ are $4 \mathmr{and} 4$

$4 \times 4 = 16 \mathmr{and} 4 + 4 = 8$

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

$= \left(4 x - 1\right) \left(4 x - 1\right)$