Find the Roots (Zeros) f(x)=(2x^2-8x+6)/(x^2-4x+4)

$f (x) = \frac{2 x^{2} - 8 x + 6}{x^{2} - 4 x + 4}$

Step 1

Set $\frac{2 x^{2} - 8 x + 6}{x^{2} - 4 x + 4}$ equal to $0$ .

$\frac{2 x^{2} - 8 x + 6}{x^{2} - 4 x + 4} = 0$

Step 2

Solve for $x$ .

Set the numerator equal to zero.

$2 x^{2} - 8 x + 6 = 0$

Solve the equation for $x$ .

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 8 x + 6$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) - 8 x + 6 = 0$

Factor $2$ out of $- 8 x$ .

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

Factor $2$ out of $6$ .

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

Factor $2$ out of $2 x^{2} + 2 (- 4 x)$ .

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

Factor $2$ out of $2 (x^{2} - 4 x) + 2 \cdot 3$ .

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

Factor.

Factor $x^{2} - 4 x + 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$2 ((x - 3) (x - 1)) = 0$

Remove unnecessary parentheses.

$2 (x - 3) (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x - 1 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $2 (x - 3) (x - 1) = 0$ true.

$x = 3, 1$

Step 3

Find the Roots (Zeros) f(x)=(2x^2-8x+6)/(x^2-4x+4)

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