Identify the Zeros and Their Multiplicities y=(x-1)^2(2x-3)^3

$y = {(x - 1)}^{2} {(2 x - 3)}^{3}$

Step 1

Set ${(x - 1)}^{2} {(2 x - 3)}^{3}$ equal to $0$ .

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

Step 2

Solve for $x$ .

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

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

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

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

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

Solve ${(x - 1)}^{2} = 0$ for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set ${(2 x - 3)}^{3}$ equal to $0$ and solve for $x$ .

Set ${(2 x - 3)}^{3}$ equal to $0$ .

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

Solve ${(2 x - 3)}^{3} = 0$ for $x$ .

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

$2 x - 3 = 0$

Solve 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 $1$ .

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

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

$x = 1$ (Multiplicity of $2$ )

$x = \frac{3}{2}$ (Multiplicity of $3$ )

$x = 1$ (Multiplicity of $2$ )

$x = \frac{3}{2}$ (Multiplicity of $3$ )

Step 3

Identify the Zeros and Their Multiplicities y=(x-1)^2(2x-3)^3

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