Write in Standard Form (x+3)(x-3)=x(4x+5)

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

Step 1

Subtract $x (4 x + 5)$ from both sides of the equation.

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

Step 2

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

Simplify each term.

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

Apply the distributive property.

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

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 3 (x - 3) - x (4 x + 5) = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3 - x (4 x + 5) = 0$

Combine the opposite terms in $x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3$ .

Reorder the factors in the terms $x \cdot - 3$ and $3 x$ .

$x \cdot x - 3 x + 3 x + 3 \cdot - 3 - x (4 x + 5) = 0$

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

$x \cdot x + 0 + 3 \cdot - 3 - x (4 x + 5) = 0$

Add $x \cdot x$ and $0$ .

$x \cdot x + 3 \cdot - 3 - x (4 x + 5) = 0$

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + 3 \cdot - 3 - x (4 x + 5) = 0$

Multiply $3$ by $- 3$ .

$x^{2} - 9 - x (4 x + 5) = 0$

Apply the distributive property.

$x^{2} - 9 - x (4 x) - x \cdot 5 = 0$

Rewrite using the commutative property of multiplication.

$x^{2} - 9 - 1 \cdot 4 x \cdot x - x \cdot 5 = 0$

Multiply $5$ by $- 1$ .

$x^{2} - 9 - 1 \cdot 4 x \cdot x - 5 x = 0$

Simplify each term.

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

Move $x$ .

$x^{2} - 9 - 1 \cdot 4 (x \cdot x) - 5 x = 0$

Multiply $x$ by $x$ .

$x^{2} - 9 - 1 \cdot 4 x^{2} - 5 x = 0$

Multiply $- 1$ by $4$ .

$x^{2} - 9 - 4 x^{2} - 5 x = 0$

Subtract $4 x^{2}$ from $x^{2}$ .

$- 3 x^{2} - 9 - 5 x = 0$

Step 3

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$a x^{2} + b x + c = 0$

Step 4

Move $- 9$ .

$- 3 x^{2} - 5 x - 9 = 0$

Step 5

Write in Standard Form (x+3)(x-3)=x(4x+5)

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