Find the Roots (Zeros) x^4-8x^3+24x^2-32x+16=0

$x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 16 = 0$

Step 1

Factor the left side of the equation.

Make each term match the terms from the binomial theorem formula.

$x^{4} - 4 \cdot (2 x^{3}) + 6 \cdot (2^{2} x^{2}) - 4 \cdot (2^{3} x) + 2^{4} = 0$

Factor $x^{4} - 4 \cdot 2 x^{3} + 6 \cdot 2^{2} x^{2} - 4 \cdot 2^{3} x + 2^{4}$ using the binomial theorem.

${(2 - x)}^{4} = 0$

Step 2

Set the $2 - x$ equal to $0$ .

$2 - x = 0$

Step 3

Solve for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

Step 4