Identify the Zeros and Their Multiplicities f(x)=3x^4-2x^3+74x^2-50x-25

$f (x) = 3 x^{4} - 2 x^{3} + 74 x^{2} - 50 x - 25$

Step 1

Set $3 x^{4} - 2 x^{3} + 74 x^{2} - 50 x - 25$ equal to $0$ .

$3 x^{4} - 2 x^{3} + 74 x^{2} - 50 x - 25 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$- 2 x^{3} - 50 x + 3 x^{4} + 74 x^{2} - 25 = 0$

Factor $- 2 x$ out of $- 2 x^{3} - 50 x$ .

Factor $- 2 x$ out of $- 2 x^{3}$ .

$- 2 x \cdot x^{2} - 50 x + 3 x^{4} + 74 x^{2} - 25 = 0$

Factor $- 2 x$ out of $- 50 x$ .

$- 2 x \cdot x^{2} - 2 x \cdot 25 + 3 x^{4} + 74 x^{2} - 25 = 0$

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

$- 2 x (x^{2} + 25) + 3 x^{4} + 74 x^{2} - 25 = 0$

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

$- 2 x (x^{2} + 25) + 3 {(x^{2})}^{2} + 74 x^{2} - 25 = 0$

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

$- 2 x (x^{2} + 25) + 3 u^{2} + 74 u - 25 = 0$

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 $a \cdot c = 3 \cdot - 25 = - 75$ and whose sum is $b = 74$ .

Factor $74$ out of $74 u$ .

$- 2 x (x^{2} + 25) + 3 u^{2} + 74 (u) - 25 = 0$

Rewrite $74$ as $- 1$ plus $75$

$- 2 x (x^{2} + 25) + 3 u^{2} + (- 1 + 75) u - 25 = 0$

Apply the distributive property.

$- 2 x (x^{2} + 25) + 3 u^{2} - 1 u + 75 u - 25 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 2 x (x^{2} + 25) + 3 u^{2} - 1 u + 75 u - 25 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 u - 1$ .

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

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

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

Factor $x^{2} + 25$ out of $- 2 x (x^{2} + 25) + (3 x^{2} - 1) (x^{2} + 25)$ .

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

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

Factor $x^{2} + 25$ out of $(3 x^{2} - 1) (x^{2} + 25)$ .

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

Factor $x^{2} + 25$ out of $(x^{2} + 25) (- 2 x) + (x^{2} + 25) (3 x^{2} - 1)$ .

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

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

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

Factor by grouping.

Reorder terms.

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

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 $a \cdot c = 3 \cdot - 1 = - 3$ and whose sum is $b = - 2$ .

Factor $- 2$ out of $- 2 u$ .

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

Rewrite $- 2$ as $1$ plus $- 3$

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

Apply the distributive property.

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

Multiply $u$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $3 u + 1$ .

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

Factor.

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

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

Remove unnecessary parentheses.

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

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

$x^{2} + 25 = 0$

$3 x + 1 = 0$

$x - 1 = 0$

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 specified 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.

$x = 5 i, - 5 i$

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

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.

$x = 1$

The final solution is all the values that make $(x^{2} + 25) (3 x + 1) (x - 1) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 5 i$ (Multiplicity of $1$ )

$x = - 5 i$ (Multiplicity of $1$ )

$x = - \frac{1}{3}$ (Multiplicity of $1$ )

$x = 1$ (Multiplicity of $1$ )

$x = 5 i$ (Multiplicity of $1$ )

$x = - 5 i$ (Multiplicity of $1$ )

$x = - \frac{1}{3}$ (Multiplicity of $1$ )

$x = 1$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=3x^4-2x^3+74x^2-50x-25

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