Find the Roots (Zeros) 2x^4-5x^3+53x^2-125x+75=0

$2 x^{4} - 5 x^{3} + 53 x^{2} - 125 x + 75 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 5 x^{3} - 125 x + 2 x^{4} + 53 x^{2} + 75 = 0$

Factor $- 5 x$ out of $- 5 x^{3} - 125 x$ .

Factor $- 5 x$ out of $- 5 x^{3}$ .

$- 5 x \cdot x^{2} - 125 x + 2 x^{4} + 53 x^{2} + 75 = 0$

Factor $- 5 x$ out of $- 125 x$ .

$- 5 x \cdot x^{2} - 5 x \cdot 25 + 2 x^{4} + 53 x^{2} + 75 = 0$

Factor $- 5 x$ out of $- 5 x (x^{2}) - 5 x (25)$ .

$- 5 x (x^{2} + 25) + 2 x^{4} + 53 x^{2} + 75 = 0$

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

$- 5 x (x^{2} + 25) + 2 {(x^{2})}^{2} + 53 x^{2} + 75 = 0$

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

$- 5 x (x^{2} + 25) + 2 u^{2} + 53 u + 75 = 0$

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 $a \cdot c = 2 \cdot 75 = 150$ and whose sum is $b = 53$ .

Factor $53$ out of $53 u$ .

$- 5 x (x^{2} + 25) + 2 u^{2} + 53 (u) + 75 = 0$

Rewrite $53$ as $3$ plus $50$

$- 5 x (x^{2} + 25) + 2 u^{2} + (3 + 50) u + 75 = 0$

Apply the distributive property.

$- 5 x (x^{2} + 25) + 2 u^{2} + 3 u + 50 u + 75 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 5 x (x^{2} + 25) + 2 u^{2} + 3 u + 50 u + 75 = 0$

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

$- 5 x (x^{2} + 25) + u (2 u + 3) + 25 (2 u + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u + 3$ .

$- 5 x (x^{2} + 25) + (2 u + 3) (u + 25) = 0$

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

$- 5 x (x^{2} + 25) + (2 x^{2} + 3) (x^{2} + 25) = 0$

Factor $x^{2} + 25$ out of $- 5 x (x^{2} + 25) + (2 x^{2} + 3) (x^{2} + 25)$ .

Factor $x^{2} + 25$ out of $- 5 x (x^{2} + 25)$ .

$(x^{2} + 25) (- 5 x) + (2 x^{2} + 3) (x^{2} + 25) = 0$

Factor $x^{2} + 25$ out of $(2 x^{2} + 3) (x^{2} + 25)$ .

$(x^{2} + 25) (- 5 x) + (x^{2} + 25) (2 x^{2} + 3) = 0$

Factor $x^{2} + 25$ out of $(x^{2} + 25) (- 5 x) + (x^{2} + 25) (2 x^{2} + 3)$ .

$(x^{2} + 25) (- 5 x + 2 x^{2} + 3) = 0$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x^{2} + 25) (- 5 u + 2 u^{2} + 3) = 0$

Factor by grouping.

Reorder terms.

$(x^{2} + 25) (2 u^{2} - 5 u + 3) = 0$

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 u$ .

$(x^{2} + 25) (2 u^{2} - 5 u + 3) = 0$

Rewrite $- 5$ as $- 2$ plus $- 3$

$(x^{2} + 25) (2 u^{2} + (- 2 - 3) u + 3) = 0$

Apply the distributive property.

$(x^{2} + 25) (2 u^{2} - 2 u - 3 u + 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{2} + 25) ((2 u^{2} - 2 u) - 3 u + 3) = 0$

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

$(x^{2} + 25) (2 u (u - 1) - 3 (u - 1)) = 0$

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

$(x^{2} + 25) ((u - 1) (2 u - 3)) = 0$

Factor.

Replace all occurrences of $u$ with $x$ .

$(x^{2} + 25) ((x - 1) (2 x - 3)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 25) (x - 1) (2 x - 3) = 0$

Step 2

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

$x^{2} + 25 = 0$

$x - 1 = 0$

$2 x - 3 = 0$

Step 3

Set $x^{2} + 25$ equal to $0$ and solve for $x$ .

Set $x^{2} + 25$ equal to $0$ .

$x^{2} + 25 = 0$

Solve $x^{2} + 25 = 0$ for $x$ .

Subtract $25$ from both sides of the equation.

$x^{2} = - 25$

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

$x = \pm \sqrt{- 25}$

Simplify $\pm \sqrt{- 25}$ .

Rewrite $- 25$ as $- 1 (25)$ .

$x = \pm \sqrt{- 1 (25)}$

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{25}$

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

$x = \pm i \cdot \sqrt{25}$

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

$x = \pm i \cdot \sqrt{5^{2}}$

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

$x = \pm i \cdot 5$

Move $5$ to the left of $i$ .

$x = \pm 5 i$

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.

$x = 5 i$

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

$x = - 5 i$

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

$x = 5 i, - 5 i$

Step 4

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 5

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 6

The final solution is all the values that make $(x^{2} + 25) (x - 1) (2 x - 3) = 0$ true.

$x = 5 i, - 5 i, 1, \frac{3}{2}$

Step 7

Find the Roots (Zeros) 2x^4-5x^3+53x^2-125x+75=0

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