Solve (1)(3x+2)^2=16

$$( 1 ) ( 3 x + 2 ) ^ { 2 } = 16$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=-2$$

Steps Using Factoring By Grouping

Divide both sides by 1.

$$\left(3x+2\right)^{2}=16$$

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

$$9x^{2}+12x+4=16$$

Subtract 16 from both sides.

$$9x^{2}+12x+4-16=0$$

Subtract 16 from 4 to get -12.

$$9x^{2}+12x-12=0$$

Divide both sides by 3.

$$3x^{2}+4x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\left(-4\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite 3x^{2}+4x-4 as \left(3x^{2}-2x\right)+\left(6x-4\right).

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

Factor out x in the first and 2 in the second group.

$$x\left(3x-2\right)+2\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(x+2\right)$$

To find equation solutions, solve 3x-2=0 and x+2=0.

$$x=\frac{2}{3} x=-2$$

Steps Using the Quadratic Formula

Divide both sides by 1.

$$\left(3x+2\right)^{2}=16$$

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

$$9x^{2}+12x+4=16$$

Subtract 16 from both sides.

$$9x^{2}+12x+4-16=0$$

Subtract 16 from 4 to get -12.

$$9x^{2}+12x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 12 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 9\left(-12\right)}}{2\times 9}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 9\left(-12\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-12±\sqrt{144-36\left(-12\right)}}{2\times 9}$$

Multiply -36 times -12.

$$x=\frac{-12±\sqrt{144+432}}{2\times 9}$$

Add 144 to 432.

$$x=\frac{-12±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$x=\frac{-12±24}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-12±24}{18}$$

Now solve the equation x=\frac{-12±24}{18} when ± is plus. Add -12 to 24.

$$x=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{-12±24}{18} when ± is minus. Subtract 24 from -12.

$$x=-\frac{36}{18}$$

Divide -36 by 18.

$$x=-2$$

The equation is now solved.

$$x=\frac{2}{3} x=-2$$

Steps for Completing the Square

Divide both sides by 1.

$$\left(3x+2\right)^{2}=16$$

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

$$9x^{2}+12x+4=16$$

Subtract 4 from both sides.

$$9x^{2}+12x=16-4$$

Subtract 4 from 16 to get 12.

$$9x^{2}+12x=12$$

Divide both sides by 9.

$$\frac{9x^{2}+12x}{9}=\frac{12}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{12}{9}x=\frac{12}{9}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{4}{3}x=\frac{12}{9}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{4}{3}x=\frac{4}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{4}{3}+\frac{4}{9}$$

Add \frac{4}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{16}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x+\frac{2}{3}=\frac{4}{3} x+\frac{2}{3}=-\frac{4}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

$$x=\frac{2}{3} x=-2$$