Find Where Undefined/Discontinuous (x^4-2x^3+x^2)/(14x^4-8x^3-6x^2)

$\frac{(x^{4} - 2 x^{3} + x^{2})}{14 x^{4} - 8 x^{3} - 6 x^{2}}$

Step 1

Set the denominator in $\frac{x^{4} - 2 x^{3} + x^{2}}{14 x^{4} - 8 x^{3} - 6 x^{2}}$ equal to $0$ to find where the expression is undefined.

$14 x^{4} - 8 x^{3} - 6 x^{2} = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2 x^{2}$ out of $14 x^{4} - 8 x^{3} - 6 x^{2}$ .

Factor $2 x^{2}$ out of $14 x^{4}$ .

$2 x^{2} (7 x^{2}) - 8 x^{3} - 6 x^{2} = 0$

Factor $2 x^{2}$ out of $- 8 x^{3}$ .

$2 x^{2} (7 x^{2}) + 2 x^{2} (- 4 x) - 6 x^{2} = 0$

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

$2 x^{2} (7 x^{2}) + 2 x^{2} (- 4 x) + 2 x^{2} (- 3) = 0$

Factor $2 x^{2}$ out of $2 x^{2} (7 x^{2}) + 2 x^{2} (- 4 x)$ .

$2 x^{2} (7 x^{2} - 4 x) + 2 x^{2} (- 3) = 0$

Factor $2 x^{2}$ out of $2 x^{2} (7 x^{2} - 4 x) + 2 x^{2} (- 3)$ .

$2 x^{2} (7 x^{2} - 4 x - 3) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 7 \cdot - 3 = - 21$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 x$ .

$2 x^{2} (7 x^{2} - 4 x - 3) = 0$

Rewrite $- 4$ as $3$ plus $- 7$

$2 x^{2} (7 x^{2} + (3 - 7) x - 3) = 0$

Apply the distributive property.

$2 x^{2} (7 x^{2} + 3 x - 7 x - 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 x^{2} ((7 x^{2} + 3 x) - 7 x - 3) = 0$

Factor out the greatest common factor (GCF) from each group.

$2 x^{2} (x (7 x + 3) - (7 x + 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 x + 3$ .

$2 x^{2} ((7 x + 3) (x - 1)) = 0$

Remove unnecessary parentheses.

$2 x^{2} (7 x + 3) (x - 1) = 0$

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

$x^{2} = 0$

$7 x + 3 = 0$

$x - 1 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $7 x + 3$ equal to $0$ and solve for $x$ .

Set $7 x + 3$ equal to $0$ .

$7 x + 3 = 0$

Solve $7 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$7 x = - 3$

Divide each term in $7 x = - 3$ by $7$ and simplify.

Divide each term in $7 x = - 3$ by $7$ .

$\frac{7 x}{7} = \frac{- 3}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 3}{7}$

Divide $x$ by $1$ .

$x = \frac{- 3}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{7}$

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 $2 x^{2} (7 x + 3) (x - 1) = 0$ true.

$x = 0, - \frac{3}{7}, 1$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - \frac{3}{7}, x = 0, x = 1$

Step 4

Find Where Undefined/Discontinuous (x^4-2x^3+x^2)/(14x^4-8x^3-6x^2)

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