Solve by Factoring 16x^4+47x^2=3

$16 x^{4} + 47 x^{2} = 3$

Step 1

Subtract $3$ from both sides of the equation.

$16 x^{4} + 47 x^{2} - 3 = 0$

Step 2

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$16 {(x^{2})}^{2} + 47 x^{2} - 3 = 0$

Step 3

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$16 u^{2} + 47 u - 3 = 0$

Step 4

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 $a \cdot c = 16 \cdot - 3 = - 48$ and whose sum is $b = 47$ .

Factor $47$ out of $47 u$ .

$16 u^{2} + 47 (u) - 3 = 0$

Rewrite $47$ as $- 1$ plus $48$

$16 u^{2} + (- 1 + 48) u - 3 = 0$

Apply the distributive property.

$16 u^{2} - 1 u + 48 u - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(16 u^{2} - 1 u) + 48 u - 3 = 0$

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

$u (16 u - 1) + 3 (16 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $16 u - 1$ .

$(16 u - 1) (u + 3) = 0$

Step 5

Replace all occurrences of $u$ with $x^{2}$ .

$(16 x^{2} - 1) (x^{2} + 3) = 0$

Step 6

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

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

Step 7

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

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

Step 8

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 x$ and $b = 1$ .

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

Step 9

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

$4 x + 1 = 0$

$4 x - 1 = 0$

$x^{2} + 3 = 0$

Step 10

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 11

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 $x$ .

Add $1$ to 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 $1$ .

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

Step 12

Set $x^{2} + 3$ equal to $0$ and solve for $x$ .

Set $x^{2} + 3$ equal to $0$ .

$x^{2} + 3 = 0$

Solve $x^{2} + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

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

$x = \pm \sqrt{- 3}$

Simplify $\pm \sqrt{- 3}$ .

Rewrite $- 3$ as $- 1 (3)$ .

$x = \pm \sqrt{- 1 (3)}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{3}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = i \sqrt{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = i \sqrt{3}, - i \sqrt{3}$

Step 13

The final solution is all the values that make $(4 x + 1) (4 x - 1) (x^{2} + 3) = 0$ true.

$x = - \frac{1}{4}, \frac{1}{4}, i \sqrt{3}, - i \sqrt{3}$

Solve by Factoring 16x^4+47x^2=3

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