Find the x and y Intercepts y=(x+2)(2x+1)(x+3)

$y = (x + 2) (2 x + 1) (x + 3)$

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = (x + 2) (2 x + 1) (x + 3)$

Solve the equation.

Rewrite the equation as $(x + 2) (2 x + 1) (x + 3) = 0$ .

$(x + 2) (2 x + 1) (x + 3) = 0$

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

$x + 2 = 0$

$2 x + 1 = 0$

$x + 3 = 0$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ 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.

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

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x + 2) (2 x + 1) (x + 3) = 0$ true.

$x = - 2, - \frac{1}{2}, - 3$

x-intercept(s) in point form.

x-intercept(s): $(- 2, 0), (- \frac{1}{2}, 0), (- 3, 0)$

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = ((0) + 2) (2 (0) + 1) ((0) + 3)$

Solve the equation.

Remove parentheses.

$y = (0 + 2) (2 (0) + 1) ((0) + 3)$

Remove parentheses.

$y = (0 + 2) (2 (0) + 1) (0 + 3)$

Remove parentheses.

$y = ((0) + 2) (2 (0) + 1) ((0) + 3)$

Simplify $((0) + 2) (2 (0) + 1) ((0) + 3)$ .

Add $0$ and $2$ .

$y = 2 (2 (0) + 1) ((0) + 3)$

Multiply $2$ by $0$ .

$y = 2 (0 + 1) ((0) + 3)$

Add $0$ and $1$ .

$y = 2 \cdot 1 ((0) + 3)$

Multiply $2$ by $1$ .

$y = 2 ((0) + 3)$

Add $0$ and $3$ .

$y = 2 \cdot 3$

Multiply $2$ by $3$ .

$y = 6$

y-intercept(s) in point form.

y-intercept(s): $(0, 6)$

List the intersections.

x-intercept(s): $(- 2, 0), (- \frac{1}{2}, 0), (- 3, 0)$

y-intercept(s): $(0, 6)$

Find the x and y Intercepts y=(x+2)(2x+1)(x+3)

Related Questions

Support LCMGCF.com by disabling your adblocker.