Solve for x x^3-7x+6=(x-1)(x+3)(x-2)

$x^{3} - 7 x + 6 = (x - 1) (x + 3) (x - 2)$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$(x - 1) (x + 3) (x - 2) = x^{3} - 7 x + 6$

Step 2

Simplify $(x - 1) (x + 3) (x - 2)$ .

Rewrite.

$0 + 0 + (x - 1) (x + 3) (x - 2) = x^{3} - 7 x + 6$

Simplify by adding zeros.

$(x - 1) (x + 3) (x - 2) = x^{3} - 7 x + 6$

Expand $(x - 1) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$(x (x + 3) - 1 (x + 3)) (x - 2) = x^{3} - 7 x + 6$

Apply the distributive property.

$(x \cdot x + x \cdot 3 - 1 (x + 3)) (x - 2) = x^{3} - 7 x + 6$

Apply the distributive property.

$(x \cdot x + x \cdot 3 - 1 x - 1 \cdot 3) (x - 2) = x^{3} - 7 x + 6$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$(x^{2} + x \cdot 3 - 1 x - 1 \cdot 3) (x - 2) = x^{3} - 7 x + 6$

Move $3$ to the left of $x$ .

$(x^{2} + 3 \cdot x - 1 x - 1 \cdot 3) (x - 2) = x^{3} - 7 x + 6$

Rewrite $- 1 x$ as $- x$ .

$(x^{2} + 3 x - x - 1 \cdot 3) (x - 2) = x^{3} - 7 x + 6$

Multiply $- 1$ by $3$ .

$(x^{2} + 3 x - x - 3) (x - 2) = x^{3} - 7 x + 6$

Subtract $x$ from $3 x$ .

$(x^{2} + 2 x - 3) (x - 2) = x^{3} - 7 x + 6$

Expand $(x^{2} + 2 x - 3) (x - 2)$ by multiplying each term in the first expression by each term in the second expression.

$x^{2} x + x^{2} \cdot - 2 + 2 x \cdot x + 2 x \cdot - 2 - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

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 $1$ .

$x^{2} x^{1} + x^{2} \cdot - 2 + 2 x \cdot x + 2 x \cdot - 2 - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

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

$x^{2 + 1} + x^{2} \cdot - 2 + 2 x \cdot x + 2 x \cdot - 2 - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

Add $2$ and $1$ .

$x^{3} + x^{2} \cdot - 2 + 2 x \cdot x + 2 x \cdot - 2 - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

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

$x^{3} - 2 \cdot x^{2} + 2 x \cdot x + 2 x \cdot - 2 - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

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

Move $x$ .

$x^{3} - 2 x^{2} + 2 (x \cdot x) + 2 x \cdot - 2 - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

Multiply $x$ by $x$ .

$x^{3} - 2 x^{2} + 2 x^{2} + 2 x \cdot - 2 - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

Multiply $- 2$ by $2$ .

$x^{3} - 2 x^{2} + 2 x^{2} - 4 x - 3 x - 3 \cdot - 2 = x^{3} - 7 x + 6$

Multiply $- 3$ by $- 2$ .

$x^{3} - 2 x^{2} + 2 x^{2} - 4 x - 3 x + 6 = x^{3} - 7 x + 6$

Simplify by adding terms.

Combine the opposite terms in $x^{3} - 2 x^{2} + 2 x^{2} - 4 x - 3 x + 6$ .

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

$x^{3} + 0 - 4 x - 3 x + 6 = x^{3} - 7 x + 6$

Add $x^{3}$ and $0$ .

$x^{3} - 4 x - 3 x + 6 = x^{3} - 7 x + 6$

Subtract $3 x$ from $- 4 x$ .

$x^{3} - 7 x + 6 = x^{3} - 7 x + 6$

Step 3

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

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

$x^{3} - 7 x + 6 - x^{3} = - 7 x + 6$

Add $7 x$ to both sides of the equation.

$x^{3} - 7 x + 6 - x^{3} + 7 x = 6$

Combine the opposite terms in $x^{3} - 7 x + 6 - x^{3} + 7 x$ .

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

$- 7 x + 6 + 0 + 7 x = 6$

Add $- 7 x + 6$ and $0$ .

$- 7 x + 6 + 7 x = 6$

Add $- 7 x$ and $7 x$ .

$0 + 6 = 6$

Add $0$ and $6$ .

$6 = 6$

Step 4

Since $6 = 6$ , the equation will always be true for any value of $x$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x x^3-7x+6=(x-1)(x+3)(x-2)

Related Questions

Support LCMGCF.com by disabling your adblocker.