Find the x and y Intercepts P(x)=x^3+5x^2-14x

$P (x) = x^{3} + 5 x^{2} - 14 x$

Find the x-intercepts.

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

$0 = x^{3} + 5 x^{2} - 14 x$

Solve the equation.

Rewrite the equation as $x^{3} + 5 x^{2} - 14 x = 0$ .

$x^{3} + 5 x^{2} - 14 x = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{3} + 5 x^{2} - 14 x$ .

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

$x \cdot x^{2} + 5 x^{2} - 14 x = 0$

Factor $x$ out of $5 x^{2}$ .

$x \cdot x^{2} + x (5 x) - 14 x = 0$

Factor $x$ out of $- 14 x$ .

$x \cdot x^{2} + x (5 x) + x \cdot - 14 = 0$

Factor $x$ out of $x \cdot x^{2} + x (5 x)$ .

$x (x^{2} + 5 x) + x \cdot - 14 = 0$

Factor $x$ out of $x (x^{2} + 5 x) + x \cdot - 14$ .

$x (x^{2} + 5 x - 14) = 0$

Factor.

Factor $x^{2} + 5 x - 14$ 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 $- 14$ and whose sum is $5$ .

$- 2, 7$

Write the factored form using these integers.

$x ((x - 2) (x + 7)) = 0$

Remove unnecessary parentheses.

$x (x - 2) (x + 7) = 0$

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

$x = 0$

$x - 2 = 0$

$x + 7 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = 2$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $x (x - 2) (x + 7) = 0$ true.

$x = 0, 2, - 7$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (2, 0), (- 7, 0)$

Find the y-intercepts.

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

$y = {(0)}^{3} + 5 {(0)}^{2} - 14 \cdot 0$

Solve the equation.

Remove parentheses.

$y = 0^{3} + 5 {(0)}^{2} - 14 \cdot 0$

Remove parentheses.

$y = 0^{3} + 5 \cdot 0^{2} - 14 \cdot 0$

Remove parentheses.

$y = {(0)}^{3} + 5 {(0)}^{2} - 14 \cdot 0$

Simplify ${(0)}^{3} + 5 {(0)}^{2} - 14 \cdot 0$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 + 5 {(0)}^{2} - 14 \cdot 0$

Raising $0$ to any positive power yields $0$ .

$y = 0 + 5 \cdot 0 - 14 \cdot 0$

Multiply $5$ by $0$ .

$y = 0 + 0 - 14 \cdot 0$

Multiply $- 14$ by $0$ .

$y = 0 + 0 + 0$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

List the intersections.

x-intercept(s): $(0, 0), (2, 0), (- 7, 0)$

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

Find the x and y Intercepts P(x)=x^3+5x^2-14x

Related Questions

Support LCMGCF.com by disabling your adblocker.