Write in Standard Form (x-4)^2+8=0

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

Step 1

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

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

Step 2

Simplify each term.

Rewrite ${(x - 4)}^{2}$ as $(x - 4) (x - 4)$ .

$(x - 4) (x - 4) + 8 = 0$

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

Apply the distributive property.

$x (x - 4) - 4 (x - 4) + 8 = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 4 - 4 (x - 4) + 8 = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 4 - 4 x - 4 \cdot - 4 + 8 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 4 - 4 x - 4 \cdot - 4 + 8 = 0$

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

$x^{2} - 4 \cdot x - 4 x - 4 \cdot - 4 + 8 = 0$

Multiply $- 4$ by $- 4$ .

$x^{2} - 4 x - 4 x + 16 + 8 = 0$

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

$x^{2} - 8 x + 16 + 8 = 0$

Step 3

Add $16$ and $8$ .

$x^{2} - 8 x + 24 = 0$

Step 4