Find the Roots (Zeros) y=3x(x-4)^3(x+6)^2(2x+1)^4

$y = 3 x {(x - 4)}^{3} {(x + 6)}^{2} {(2 x + 1)}^{4}$

Set $3 x {(x - 4)}^{3} {(x + 6)}^{2} {(2 x + 1)}^{4}$ equal to $0$ .

$3 x {(x - 4)}^{3} {(x + 6)}^{2} {(2 x + 1)}^{4} = 0$

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 $0$ .

$x = 0$

${(x - 4)}^{3} = 0$

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

${(2 x + 1)}^{4} = 0$

Set $x$ equal to $0$ .

$x = 0$

Set ${(x - 4)}^{3}$ equal to $0$ and solve for $x$ .

Set ${(x - 4)}^{3}$ equal to $0$ .

${(x - 4)}^{3} = 0$

Solve ${(x - 4)}^{3} = 0$ for $x$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set ${(x + 6)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 6)}^{2}$ equal to $0$ .

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

Solve ${(x + 6)}^{2} = 0$ for $x$ .

Set the $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Set ${(2 x + 1)}^{4}$ equal to $0$ and solve for $x$ .

Set ${(2 x + 1)}^{4}$ equal to $0$ .

${(2 x + 1)}^{4} = 0$

Solve ${(2 x + 1)}^{4} = 0$ for $x$ .

Set the $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

The final solution is all the values that make $3 x {(x - 4)}^{3} {(x + 6)}^{2} {(2 x + 1)}^{4} = 0$ true.

$x = 0, 4, - 6, - \frac{1}{2}$

Find the Roots (Zeros) y=3x(x-4)^3(x+6)^2(2x+1)^4

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