Solve for x (x+1)(2x+5)=(2x+3)(x-4)+5

$(x + 1) (2 x + 5) = (2 x + 3) (x - 4) + 5$

Step 1

Simplify $(x + 1) (2 x + 5)$ .

Rewrite.

$0 + 0 + (x + 1) (2 x + 5) = (2 x + 3) (x - 4) + 5$

Simplify by adding zeros.

$(x + 1) (2 x + 5) = (2 x + 3) (x - 4) + 5$

Expand $(x + 1) (2 x + 5)$ using the FOIL Method.

Apply the distributive property.

$x (2 x + 5) + 1 (2 x + 5) = (2 x + 3) (x - 4) + 5$

Apply the distributive property.

$x (2 x) + x \cdot 5 + 1 (2 x + 5) = (2 x + 3) (x - 4) + 5$

Apply the distributive property.

$x (2 x) + x \cdot 5 + 1 (2 x) + 1 \cdot 5 = (2 x + 3) (x - 4) + 5$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot 5 + 1 (2 x) + 1 \cdot 5 = (2 x + 3) (x - 4) + 5$

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

Move $x$ .

$2 (x \cdot x) + x \cdot 5 + 1 (2 x) + 1 \cdot 5 = (2 x + 3) (x - 4) + 5$

Multiply $x$ by $x$ .

$2 x^{2} + x \cdot 5 + 1 (2 x) + 1 \cdot 5 = (2 x + 3) (x - 4) + 5$

Move $5$ to the left of $x$ .

$2 x^{2} + 5 \cdot x + 1 (2 x) + 1 \cdot 5 = (2 x + 3) (x - 4) + 5$

Multiply $2 x$ by $1$ .

$2 x^{2} + 5 x + 2 x + 1 \cdot 5 = (2 x + 3) (x - 4) + 5$

Multiply $5$ by $1$ .

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

Add $5 x$ and $2 x$ .

$2 x^{2} + 7 x + 5 = (2 x + 3) (x - 4) + 5$

Step 2

Simplify $(2 x + 3) (x - 4) + 5$ .

Simplify each term.

Expand $(2 x + 3) (x - 4)$ using the FOIL Method.

Apply the distributive property.

$2 x^{2} + 7 x + 5 = 2 x (x - 4) + 3 (x - 4) + 5$

Apply the distributive property.

$2 x^{2} + 7 x + 5 = 2 x \cdot x + 2 x \cdot - 4 + 3 (x - 4) + 5$

Apply the distributive property.

$2 x^{2} + 7 x + 5 = 2 x \cdot x + 2 x \cdot - 4 + 3 x + 3 \cdot - 4 + 5$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$2 x^{2} + 7 x + 5 = 2 (x \cdot x) + 2 x \cdot - 4 + 3 x + 3 \cdot - 4 + 5$

Multiply $x$ by $x$ .

$2 x^{2} + 7 x + 5 = 2 x^{2} + 2 x \cdot - 4 + 3 x + 3 \cdot - 4 + 5$

Multiply $- 4$ by $2$ .

$2 x^{2} + 7 x + 5 = 2 x^{2} - 8 x + 3 x + 3 \cdot - 4 + 5$

Multiply $3$ by $- 4$ .

$2 x^{2} + 7 x + 5 = 2 x^{2} - 8 x + 3 x - 12 + 5$

Add $- 8 x$ and $3 x$ .

$2 x^{2} + 7 x + 5 = 2 x^{2} - 5 x - 12 + 5$

Add $- 12$ and $5$ .

$2 x^{2} + 7 x + 5 = 2 x^{2} - 5 x - 7$

Step 3

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

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

$2 x^{2} + 7 x + 5 - 2 x^{2} = - 5 x - 7$

Add $5 x$ to both sides of the equation.

$2 x^{2} + 7 x + 5 - 2 x^{2} + 5 x = - 7$

Combine the opposite terms in $2 x^{2} + 7 x + 5 - 2 x^{2} + 5 x$ .

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

$7 x + 5 + 0 + 5 x = - 7$

Add $7 x + 5$ and $0$ .

$7 x + 5 + 5 x = - 7$

Add $7 x$ and $5 x$ .

$12 x + 5 = - 7$

Step 4

Move all terms not containing $x$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$12 x = - 7 - 5$

Subtract $5$ from $- 7$ .

$12 x = - 12$

Step 5

Divide each term in $12 x = - 12$ by $12$ and simplify.

Divide each term in $12 x = - 12$ by $12$ .

$\frac{12 x}{12} = \frac{- 12}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 x}{12} = \frac{- 12}{12}$

Divide $x$ by $1$ .

$x = \frac{- 12}{12}$

Simplify the right side.

Divide $- 12$ by $12$ .

$x = - 1$

Solve for x (x+1)(2x+5)=(2x+3)(x-4)+5

Related Questions

Support LCMGCF.com by disabling your adblocker.