Find the Roots (Zeros) x^3+x^2-22x-40=(x+2)(x+4)(x-5)

$x^{3} + x^{2} - 22 x - 40 = (x + 2) (x + 4) (x - 5)$

Step 1

Simplify $(x + 2) (x + 4) (x - 5)$ .

Rewrite.

$x^{3} + x^{2} - 22 x - 40 = 0 + 0 + (x + 2) (x + 4) (x - 5)$

Simplify by adding zeros.

$x^{3} + x^{2} - 22 x - 40 = (x + 2) (x + 4) (x - 5)$

Expand $(x + 2) (x + 4)$ using the FOIL Method.

Apply the distributive property.

$x^{3} + x^{2} - 22 x - 40 = (x (x + 4) + 2 (x + 4)) (x - 5)$

Apply the distributive property.

$x^{3} + x^{2} - 22 x - 40 = (x \cdot x + x \cdot 4 + 2 (x + 4)) (x - 5)$

Apply the distributive property.

$x^{3} + x^{2} - 22 x - 40 = (x \cdot x + x \cdot 4 + 2 x + 2 \cdot 4) (x - 5)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{3} + x^{2} - 22 x - 40 = (x^{2} + x \cdot 4 + 2 x + 2 \cdot 4) (x - 5)$

Move $4$ to the left of $x$ .

$x^{3} + x^{2} - 22 x - 40 = (x^{2} + 4 \cdot x + 2 x + 2 \cdot 4) (x - 5)$

Multiply $2$ by $4$ .

$x^{3} + x^{2} - 22 x - 40 = (x^{2} + 4 x + 2 x + 8) (x - 5)$

Add $4 x$ and $2 x$ .

$x^{3} + x^{2} - 22 x - 40 = (x^{2} + 6 x + 8) (x - 5)$

Expand $(x^{2} + 6 x + 8) (x - 5)$ by multiplying each term in the first expression by each term in the second expression.

$x^{3} + x^{2} - 22 x - 40 = x^{2} x + x^{2} \cdot - 5 + 6 x \cdot x + 6 x \cdot - 5 + 8 x + 8 \cdot - 5$

Simplify terms.

Simplify each term.

Multiply $x^{2}$ by $x$ by adding the exponents.

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

$x^{3} + x^{2} - 22 x - 40 = x^{2} x^{1} + x^{2} \cdot - 5 + 6 x \cdot x + 6 x \cdot - 5 + 8 x + 8 \cdot - 5$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{3} + x^{2} - 22 x - 40 = x^{2 + 1} + x^{2} \cdot - 5 + 6 x \cdot x + 6 x \cdot - 5 + 8 x + 8 \cdot - 5$

Add $2$ and $1$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} + x^{2} \cdot - 5 + 6 x \cdot x + 6 x \cdot - 5 + 8 x + 8 \cdot - 5$

Move $- 5$ to the left of $x^{2}$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} - 5 \cdot x^{2} + 6 x \cdot x + 6 x \cdot - 5 + 8 x + 8 \cdot - 5$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} - 5 x^{2} + 6 (x \cdot x) + 6 x \cdot - 5 + 8 x + 8 \cdot - 5$

Multiply $x$ by $x$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} - 5 x^{2} + 6 x^{2} + 6 x \cdot - 5 + 8 x + 8 \cdot - 5$

Multiply $- 5$ by $6$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} - 5 x^{2} + 6 x^{2} - 30 x + 8 x + 8 \cdot - 5$

Multiply $8$ by $- 5$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} - 5 x^{2} + 6 x^{2} - 30 x + 8 x - 40$

Simplify by adding terms.

Add $- 5 x^{2}$ and $6 x^{2}$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} + x^{2} - 30 x + 8 x - 40$

Add $- 30 x$ and $8 x$ .

$x^{3} + x^{2} - 22 x - 40 = x^{3} + x^{2} - 22 x - 40$

Step 2

Move all terms containing $x$ to the left side of the equation.

Subtract $x^{3}$ from both sides of the equation.

$x^{3} + x^{2} - 22 x - 40 - x^{3} = x^{2} - 22 x - 40$

Subtract $x^{2}$ from both sides of the equation.

$x^{3} + x^{2} - 22 x - 40 - x^{3} - x^{2} = - 22 x - 40$

Add $22 x$ to both sides of the equation.

$x^{3} + x^{2} - 22 x - 40 - x^{3} - x^{2} + 22 x = - 40$

Combine the opposite terms in $x^{3} + x^{2} - 22 x - 40 - x^{3} - x^{2} + 22 x$ .

Subtract $x^{3}$ from $x^{3}$ .

$x^{2} - 22 x - 40 + 0 - x^{2} + 22 x = - 40$

Add $x^{2} - 22 x - 40$ and $0$ .

$x^{2} - 22 x - 40 - x^{2} + 22 x = - 40$

Subtract $x^{2}$ from $x^{2}$ .

$- 22 x - 40 + 0 + 22 x = - 40$

Add $- 22 x - 40$ and $0$ .

$- 22 x - 40 + 22 x = - 40$

Add $- 22 x$ and $22 x$ .

$0 - 40 = - 40$

Subtract $40$ from $0$ .

$- 40 = - 40$

Step 3

Since $- 40 = - 40$ , the equation will always be true.

Always true

Find the Roots (Zeros) x^3+x^2-22x-40=(x+2)(x+4)(x-5)

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