Identify the Zeros and Their Multiplicities f(x)=6x^3-x^2-294x+49

$f (x) = 6 x^{3} - x^{2} - 294 x + 49$

Step 1

Set $6 x^{3} - x^{2} - 294 x + 49$ equal to $0$ .

$6 x^{3} - x^{2} - 294 x + 49 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 x^{3} - x^{2}) - 294 x + 49 = 0$

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

$x^{2} (6 x - 1) - 49 (6 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $6 x - 1$ .

$(6 x - 1) (x^{2} - 49) = 0$

Rewrite $49$ as $7^{2}$ .

$(6 x - 1) (x^{2} - 7^{2}) = 0$

Factor.

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 $b = 7$ .

$(6 x - 1) ((x + 7) (x - 7)) = 0$

Remove unnecessary parentheses.

$(6 x - 1) (x + 7) (x - 7) = 0$

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

$6 x - 1 = 0$

$x + 7 = 0$

$x - 7 = 0$

Set $6 x - 1$ equal to $0$ and solve for $x$ .

Set $6 x - 1$ equal to $0$ .

$6 x - 1 = 0$

Solve $6 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$6 x = 1$

Divide each term in $6 x = 1$ by $6$ and simplify.

Divide each term in $6 x = 1$ by $6$ .

$\frac{6 x}{6} = \frac{1}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{1}{6}$

Divide $x$ by $1$ .

$x = \frac{1}{6}$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

The final solution is all the values that make $(6 x - 1) (x + 7) (x - 7) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = \frac{1}{6}$ (Multiplicity of $1$ )

$x = - 7$ (Multiplicity of $1$ )

$x = 7$ (Multiplicity of $1$ )

$x = \frac{1}{6}$ (Multiplicity of $1$ )

$x = - 7$ (Multiplicity of $1$ )

$x = 7$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=6x^3-x^2-294x+49

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