Algebra Examples

$4 x^{4} - 65 x^{2} + 16 = 0$

Step 1

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

$4 {(x^{2})}^{2} - 65 x^{2} + 16 = 0$

Step 2

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

$4 u^{2} - 65 u + 16 = 0$

Step 3

Factor by grouping.

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Step 3.1

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 $a \cdot c = 4 \cdot 16 = 64$ and whose sum is $b = - 65$ .

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Step 3.1.1

Factor $- 65$ out of $- 65 u$ .

$4 u^{2} - 65 u + 16 = 0$

Step 3.1.2

Rewrite $- 65$ as $- 1$ plus $- 64$

$4 u^{2} + (- 1 - 64) u + 16 = 0$

Step 3.1.3

Apply the distributive property.

$4 u^{2} - 1 u - 64 u + 16 = 0$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) - 64 u + 16 = 0$

Step 3.2.2

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

$u (4 u - 1) - 16 (4 u - 1) = 0$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$(4 u - 1) (u - 16) = 0$

Step 4

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

$(4 x^{2} - 1) (x^{2} - 16) = 0$

Step 5

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

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

Step 6

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

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

Step 7

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

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

Step 8

Rewrite $16$ as $4^{2}$ .

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

Step 9

Factor.

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Step 9.1

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 $b = 4$ .

$(2 x + 1) (2 x - 1) ((x + 4) (x - 4)) = 0$

Step 9.2

Remove unnecessary parentheses.

$(2 x + 1) (2 x - 1) (x + 4) (x - 4) = 0$

Step 10

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

$2 x + 1 = 0$

$2 x - 1 = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 11

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 11.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 11.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 11.2.2

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

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Step 11.2.2.1

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

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

Step 11.2.2.2

Simplify the left side.

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Step 11.2.2.2.1

Cancel the common factor of $2$ .

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Step 11.2.2.2.1.1

Cancel the common factor.

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

Step 11.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 11.2.2.3

Simplify the right side.

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Step 11.2.2.3.1

Move the negative in front of the fraction.

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

Step 12

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 12.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 12.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 12.2.2

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

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Step 12.2.2.1

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

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

Step 12.2.2.2

Simplify the left side.

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Step 12.2.2.2.1

Cancel the common factor of $2$ .

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Step 12.2.2.2.1.1

Cancel the common factor.

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

Step 12.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 13

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

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Step 13.1

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

$x + 4 = 0$

Step 13.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 14

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 14.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 14.2

Add $4$ to both sides of the equation.

$x = 4$

Step 15

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

$x = - \frac{1}{2}, \frac{1}{2}, - 4, 4$