Identify the Zeros and Their Multiplicities f(x)=16x^2+32x-9

$f (x) = 16 x^{2} + 32 x - 9$

Step 1

Set $16 x^{2} + 32 x - 9$ equal to $0$ .

$16 x^{2} + 32 x - 9 = 0$

Step 2

Solve for $x$ .

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 $a \cdot c = 16 \cdot - 9 = - 144$ and whose sum is $b = 32$ .

Factor $32$ out of $32 x$ .

$16 x^{2} + 32 (x) - 9 = 0$

Rewrite $32$ as $- 4$ plus $36$

$16 x^{2} + (- 4 + 36) x - 9 = 0$

Apply the distributive property.

$16 x^{2} - 4 x + 36 x - 9 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(16 x^{2} - 4 x) + 36 x - 9 = 0$

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

$4 x (4 x - 1) + 9 (4 x - 1) = 0$

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

$(4 x - 1) (4 x + 9) = 0$

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

$4 x - 1 = 0$

$4 x + 9 = 0$

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Set $4 x + 9$ equal to $0$ and solve for $x$ .

Set $4 x + 9$ equal to $0$ .

$4 x + 9 = 0$

Solve $4 x + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$4 x = - 9$

Divide each term in $4 x = - 9$ by $4$ and simplify.

Divide each term in $4 x = - 9$ by $4$ .

$\frac{4 x}{4} = \frac{- 9}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 9}{4}$

Divide $x$ by $1$ .

$x = \frac{- 9}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{9}{4}$

The final solution is all the values that make $(4 x - 1) (4 x + 9) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = \frac{1}{4}$ (Multiplicity of $1$ )

$x = - \frac{9}{4}$ (Multiplicity of $1$ )

$x = \frac{1}{4}$ (Multiplicity of $1$ )

$x = - \frac{9}{4}$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=16x^2+32x-9

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