Solve by Factoring 4x^3+x^2-4x-1=0

$4 x^{3} + x^{2} - 4 x - 1 = 0$

Step 1

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 x^{3} + x^{2}) - 4 x - 1 = 0$

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

$x^{2} (4 x + 1) - (4 x + 1) = 0$

Step 2

Factor the polynomial by factoring out the greatest common factor, $4 x + 1$ .

$(4 x + 1) (x^{2} - 1) = 0$

Step 3

Rewrite $1$ as $1^{2}$ .

$(4 x + 1) (x^{2} - 1^{2}) = 0$

Step 4

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$(4 x + 1) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(4 x + 1) (x + 1) (x - 1) = 0$

Step 5

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

$4 x + 1 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 6

Set $4 x + 1$ equal to $0$ and solve for $x$ .

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Solve $4 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term in $4 x = - 1$ by $4$ and simplify.

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Step 7

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

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.

$x = 1$

Step 9

The final solution is all the values that make $(4 x + 1) (x + 1) (x - 1) = 0$ true.

$x = - \frac{1}{4}, - 1, 1$

Solve by Factoring 4x^3+x^2-4x-1=0

Related Questions

Support LCMGCF.com by disabling your adblocker.