Find the Roots (Zeros) y=4(x+1)^3+8

$y = 4 {(x + 1)}^{3} + 8$

Step 1

Set $4 {(x + 1)}^{3} + 8$ equal to $0$ .

$4 {(x + 1)}^{3} + 8 = 0$

Step 2

Solve for $x$ .

Simplify $4 {(x + 1)}^{3} + 8$ .

Simplify each term.

Use the Binomial Theorem.

$4 (x^{3} + 3 x^{2} \cdot 1 + 3 x \cdot 1^{2} + 1^{3}) + 8 = 0$

Simplify each term.

Multiply $3$ by $1$ .

$4 (x^{3} + 3 x^{2} + 3 x \cdot 1^{2} + 1^{3}) + 8 = 0$

One to any power is one.

$4 (x^{3} + 3 x^{2} + 3 x \cdot 1 + 1^{3}) + 8 = 0$

Multiply $3$ by $1$ .

$4 (x^{3} + 3 x^{2} + 3 x + 1^{3}) + 8 = 0$

One to any power is one.

$4 (x^{3} + 3 x^{2} + 3 x + 1) + 8 = 0$

Apply the distributive property.

$4 x^{3} + 4 (3 x^{2}) + 4 (3 x) + 4 \cdot 1 + 8 = 0$

Simplify.

Multiply $3$ by $4$ .

$4 x^{3} + 12 x^{2} + 4 (3 x) + 4 \cdot 1 + 8 = 0$

Multiply $3$ by $4$ .

$4 x^{3} + 12 x^{2} + 12 x + 4 \cdot 1 + 8 = 0$

Multiply $4$ by $1$ .

$4 x^{3} + 12 x^{2} + 12 x + 4 + 8 = 0$

Add $4$ and $8$ .

$4 x^{3} + 12 x^{2} + 12 x + 12 = 0$

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx - 2.25992104$

Step 3