Solve 4(x-3)^2=16

$$4 ( x - 3 ) ^ { 2 } = 16$$

Solve for x

$$x=5$$

$$x=1$$

Steps Using Factoring

Divide both sides by 4.

$$\left(x-3\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x-3\right)^{2}=4$$

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

$$x^{2}-6x+9=4$$

Subtract 4 from both sides.

$$x^{2}-6x+9-4=0$$

Subtract 4 from 9 to get 5.

$$x^{2}-6x+5=0$$

To solve the equation, factor x^{2}-6x+5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-5\right)\left(x-1\right)$$

To find equation solutions, solve x-5=0 and x-1=0.

$$x=5 x=1$$

Steps Using Factoring By Grouping

Divide both sides by 4.

$$\left(x-3\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x-3\right)^{2}=4$$

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

$$x^{2}-6x+9=4$$

Subtract 4 from both sides.

$$x^{2}-6x+9-4=0$$

Subtract 4 from 9 to get 5.

$$x^{2}-6x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

Rewrite x^{2}-6x+5 as \left(x^{2}-5x\right)+\left(-x+5\right).

$$\left(x^{2}-5x\right)+\left(-x+5\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(x-5\right)-\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-1\right)$$

To find equation solutions, solve x-5=0 and x-1=0.

$$x=5 x=1$$

Steps Using the Quadratic Formula

Divide both sides by 4.

$$\left(x-3\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x-3\right)^{2}=4$$

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

$$x^{2}-6x+9=4$$

Subtract 4 from both sides.

$$x^{2}-6x+9-4=0$$

Subtract 4 from 9 to get 5.

$$x^{2}-6x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$x=\frac{-\left(-6\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-6\right)±4}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±4}{2}$$

Now solve the equation x=\frac{6±4}{2} when ± is plus. Add 6 to 4.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{6±4}{2} when ± is minus. Subtract 4 from 6.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=5 x=1$$

Steps for Completing the Square

Divide both sides by 4.

$$\left(x-3\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$