Evaluate 4x(5x+7)=4(5x^2+8x+4)

$4 x (5 x + 7) = 4 (5 x^{2} + 8 x + 4)$

Step 1

Simplify $4 x (5 x + 7)$ .

Start simplifying.

$0 + 0 + 4 x (5 x + 7) = 4 (5 x^{2} + 8 x + 4)$

Simplify by multiplying through.

Apply the distributive property.

$4 x (5 x) + 4 x \cdot 7 = 4 (5 x^{2} + 8 x + 4)$

Simplify the expression.

Rewrite using the commutative property of multiplication.

$4 \cdot 5 x \cdot x + 4 x \cdot 7 = 4 (5 x^{2} + 8 x + 4)$

Multiply $7$ by $4$ .

$4 \cdot 5 x \cdot x + 28 x = 4 (5 x^{2} + 8 x + 4)$

Simplify each term.

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

Move $x$ .

$4 \cdot 5 (x \cdot x) + 28 x = 4 (5 x^{2} + 8 x + 4)$

Multiply $x$ by $x$ .

$4 \cdot 5 x^{2} + 28 x = 4 (5 x^{2} + 8 x + 4)$

Multiply $4$ by $5$ .

$20 x^{2} + 28 x = 4 (5 x^{2} + 8 x + 4)$

Step 2

Simplify $4 (5 x^{2} + 8 x + 4)$ .

Apply the distributive property.

$20 x^{2} + 28 x = 4 (5 x^{2}) + 4 (8 x) + 4 \cdot 4$

Simplify.

Multiply $5$ by $4$ .

$20 x^{2} + 28 x = 20 x^{2} + 4 (8 x) + 4 \cdot 4$

Multiply $8$ by $4$ .

$20 x^{2} + 28 x = 20 x^{2} + 32 x + 4 \cdot 4$

Multiply $4$ by $4$ .

$20 x^{2} + 28 x = 20 x^{2} + 32 x + 16$

Step 3

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

Subtract $20 x^{2}$ from both sides of the equation.

$20 x^{2} + 28 x - 20 x^{2} = 32 x + 16$

Subtract $32 x$ from both sides of the equation.

$20 x^{2} + 28 x - 20 x^{2} - 32 x = 16$

Combine the opposite terms in $20 x^{2} + 28 x - 20 x^{2} - 32 x$ .

Subtract $20 x^{2}$ from $20 x^{2}$ .

$28 x + 0 - 32 x = 16$

Add $28 x$ and $0$ .

$28 x - 32 x = 16$

Subtract $32 x$ from $28 x$ .

$- 4 x = 16$

Step 4

Divide each term in $- 4 x = 16$ by $- 4$ and simplify.

Divide each term in $- 4 x = 16$ by $- 4$ .

$\frac{- 4 x}{- 4} = \frac{16}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} = \frac{16}{- 4}$

Divide $x$ by $1$ .

$x = \frac{16}{- 4}$

Simplify the right side.

Divide $16$ by $- 4$ .

$x = - 4$

Evaluate 4x(5x+7)=4(5x^2+8x+4)

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