Solve by Factoring 16x^4-41x^2+25=0

$16 x^{4} - 41 x^{2} + 25 = 0$

Step 1

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

$16 {(x^{2})}^{2} - 41 x^{2} + 25 = 0$

Step 2

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

$16 u^{2} - 41 u + 25 = 0$

Step 3

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 $a \cdot c = 16 \cdot 25 = 400$ and whose sum is $b = - 41$ .

Factor $- 41$ out of $- 41 u$ .

$16 u^{2} - 41 u + 25 = 0$

Rewrite $- 41$ as $- 16$ plus $- 25$

$16 u^{2} + (- 16 - 25) u + 25 = 0$

Apply the distributive property.

$16 u^{2} - 16 u - 25 u + 25 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(16 u^{2} - 16 u) - 25 u + 25 = 0$

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

$16 u (u - 1) - 25 (u - 1) = 0$

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

$(u - 1) (16 u - 25) = 0$

Step 4

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

$(x^{2} - 1) (16 x^{2} - 25) = 0$

Step 5

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

$(x^{2} - 1^{2}) (16 x^{2} - 25) = 0$

Step 6

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 $b = 1$ .

$(x + 1) (x - 1) (16 x^{2} - 25) = 0$

Step 7

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

$(x + 1) (x - 1) ({(4 x)}^{2} - 25) = 0$

Step 8

Rewrite $25$ as $5^{2}$ .

$(x + 1) (x - 1) ({(4 x)}^{2} - 5^{2}) = 0$

Step 9

Factor.

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 $b = 5$ .

$(x + 1) (x - 1) ((4 x + 5) (4 x - 5)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 1) (4 x + 5) (4 x - 5) = 0$

Step 10

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

$x + 1 = 0$

$x - 1 = 0$

$4 x + 5 = 0$

$4 x - 5 = 0$

Step 11

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 12

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 13

Set $4 x + 5$ equal to $0$ and solve for $x$ .

Set $4 x + 5$ equal to $0$ .

$4 x + 5 = 0$

Solve $4 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$4 x = - 5$

Divide each term in $4 x = - 5$ by $4$ and simplify.

Divide each term in $4 x = - 5$ by $4$ .

$\frac{4 x}{4} = \frac{- 5}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 5}{4}$

Divide $x$ by $1$ .

$x = \frac{- 5}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{4}$

Step 14

Set $4 x - 5$ equal to $0$ and solve for $x$ .

Set $4 x - 5$ equal to $0$ .

$4 x - 5 = 0$

Solve $4 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$4 x = 5$

Divide each term in $4 x = 5$ by $4$ and simplify.

Divide each term in $4 x = 5$ by $4$ .

$\frac{4 x}{4} = \frac{5}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{5}{4}$

Divide $x$ by $1$ .

$x = \frac{5}{4}$

Step 15

The final solution is all the values that make $(x + 1) (x - 1) (4 x + 5) (4 x - 5) = 0$ true.

$x = - 1, 1, - \frac{5}{4}, \frac{5}{4}$

Solve by Factoring 16x^4-41x^2+25=0

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