Solve 4(x-3)^2+4=8(x-3)

$$4 ( x - 3 ) ^ { 2 } + 4 = 8 ( x - 3 )$$

Solve for x

$$x=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$4\left(x^{2}-6x+9\right)+4=8\left(x-3\right)$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36+4=8\left(x-3\right)$$

Add 36 and 4 to get 40.

$$4x^{2}-24x+40=8\left(x-3\right)$$

Use the distributive property to multiply 8 by x-3.

$$4x^{2}-24x+40=8x-24$$

Subtract 8x from both sides.

$$4x^{2}-24x+40-8x=-24$$

Combine -24x and -8x to get -32x.

$$4x^{2}-32x+40=-24$$

Add 24 to both sides.

$$4x^{2}-32x+40+24=0$$

Add 40 and 24 to get 64.

$$4x^{2}-32x+64=0$$

Divide both sides by 4.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 16=16$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite x^{2}-8x+16 as \left(x^{2}-4x\right)+\left(-4x+16\right).

$$\left(x^{2}-4x\right)+\left(-4x+16\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-4\right)-4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-4\right)$$

Rewrite as a binomial square.

$$\left(x-4\right)^{2}$$

To find equation solution, solve x-4=0.

$$x=4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$4\left(x^{2}-6x+9\right)+4=8\left(x-3\right)$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36+4=8\left(x-3\right)$$

Add 36 and 4 to get 40.

$$4x^{2}-24x+40=8\left(x-3\right)$$

Use the distributive property to multiply 8 by x-3.

$$4x^{2}-24x+40=8x-24$$

Subtract 8x from both sides.

$$4x^{2}-24x+40-8x=-24$$

Combine -24x and -8x to get -32x.

$$4x^{2}-32x+40=-24$$

Add 24 to both sides.

$$4x^{2}-32x+40+24=0$$

Add 40 and 24 to get 64.

$$4x^{2}-32x+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -32 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 4\times 64}}{2\times 4}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 4\times 64}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-32\right)±\sqrt{1024-16\times 64}}{2\times 4}$$

Multiply -16 times 64.

$$x=\frac{-\left(-32\right)±\sqrt{1024-1024}}{2\times 4}$$

Add 1024 to -1024.

$$x=\frac{-\left(-32\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{-32}{2\times 4}$$

The opposite of -32 is 32.

$$x=\frac{32}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{32}{8}$$

Divide 32 by 8.

$$x=4$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$4\left(x^{2}-6x+9\right)+4=8\left(x-3\right)$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36+4=8\left(x-3\right)$$

Add 36 and 4 to get 40.

$$4x^{2}-24x+40=8\left(x-3\right)$$

Use the distributive property to multiply 8 by x-3.

$$4x^{2}-24x+40=8x-24$$

Subtract 8x from both sides.

$$4x^{2}-24x+40-8x=-24$$

Combine -24x and -8x to get -32x.

$$4x^{2}-32x+40=-24$$

Subtract 40 from both sides.

$$4x^{2}-32x=-24-40$$

Subtract 40 from -24 to get -64.

$$4x^{2}-32x=-64$$

Divide both sides by 4.

$$\frac{4x^{2}-32x}{4}=-\frac{64}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{32}{4}\right)x=-\frac{64}{4}$$

Divide -32 by 4.

$$x^{2}-8x=-\frac{64}{4}$$

Divide -64 by 4.

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

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-16+\left(-4\right)^{2}$$

Square -4.

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

Add -16 to 16.

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

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-4=0 x-4=0$$

Add 4 to both sides of the equation.

$$x=4 x=4$$

The equation is now solved. Solutions are the same.

$$x=4$$