Identify the Zeros and Their Multiplicities f(x)=4x^3-20x^2+24x

$f (x) = 4 x^{3} - 20 x^{2} + 24 x$

Step 1

Set $4 x^{3} - 20 x^{2} + 24 x$ equal to $0$ .

$4 x^{3} - 20 x^{2} + 24 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $4 x$ out of $4 x^{3} - 20 x^{2} + 24 x$ .

Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 20 x^{2} + 24 x = 0$

Factor $4 x$ out of $- 20 x^{2}$ .

$4 x (x^{2}) + 4 x (- 5 x) + 24 x = 0$

Factor $4 x$ out of $24 x$ .

$4 x (x^{2}) + 4 x (- 5 x) + 4 x (6) = 0$

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 5 x)$ .

$4 x (x^{2} - 5 x) + 4 x (6) = 0$

Factor $4 x$ out of $4 x (x^{2} - 5 x) + 4 x (6)$ .

$4 x (x^{2} - 5 x + 6) = 0$

Factor.

Factor $x^{2} - 5 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$4 x ((x - 3) (x - 2)) = 0$

Remove unnecessary parentheses.

$4 x (x - 3) (x - 2) = 0$

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

$x = 0$

$x - 3 = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $4 x (x - 3) (x - 2) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 0$ (Multiplicity of $1$ )

$x = 3$ (Multiplicity of $1$ )

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

$x = 0$ (Multiplicity of $1$ )

$x = 3$ (Multiplicity of $1$ )

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

Step 3

Identify the Zeros and Their Multiplicities f(x)=4x^3-20x^2+24x

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