Solve 6^circ(1+x)^2=726

$$6 ^ { \circ } ( 1 + x ) ^ { 2 } = 726$$

Solve for x

$$x=10$$

$$x=-12$$

Steps Using Factoring

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{726}{6}$$

Divide 726 by 6 to get 121.

$$\left(1+x\right)^{2}=121$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$1+2x+x^{2}=121$$

Subtract 121 from both sides.

$$1+2x+x^{2}-121=0$$

Subtract 121 from 1 to get -120.

$$-120+2x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+2x-120=0$$

To solve the equation, factor x^{2}+2x-120 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.

$$a+b=2 ab=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 2.

$$a=-10 b=12$$

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

$$\left(x-10\right)\left(x+12\right)$$

To find equation solutions, solve x-10=0 and x+12=0.

$$x=10 x=-12$$

Steps Using Factoring By Grouping

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{726}{6}$$

Divide 726 by 6 to get 121.

$$\left(1+x\right)^{2}=121$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$1+2x+x^{2}=121$$

Subtract 121 from both sides.

$$1+2x+x^{2}-121=0$$

Subtract 121 from 1 to get -120.

$$-120+2x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+2x-120=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-120. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-120\right)=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 2.

$$a=-10 b=12$$

Rewrite x^{2}+2x-120 as \left(x^{2}-10x\right)+\left(12x-120\right).

$$\left(x^{2}-10x\right)+\left(12x-120\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x-10\right)+12\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+12\right)$$

To find equation solutions, solve x-10=0 and x+12=0.

$$x=10 x=-12$$

Steps Using the Quadratic Formula

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{726}{6}$$

Divide 726 by 6 to get 121.

$$\left(1+x\right)^{2}=121$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$1+2x+x^{2}=121$$

Subtract 121 from both sides.

$$1+2x+x^{2}-121=0$$

Subtract 121 from 1 to get -120.

$$-120+2x+x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+2x-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-120\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-120\right)}}{2}$$

Multiply -4 times -120.

$$x=\frac{-2±\sqrt{4+480}}{2}$$

Add 4 to 480.

$$x=\frac{-2±\sqrt{484}}{2}$$

Take the square root of 484.

$$x=\frac{-2±22}{2}$$

Now solve the equation x=\frac{-2±22}{2} when ± is plus. Add -2 to 22.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{-2±22}{2} when ± is minus. Subtract 22 from -2.

$$x=-\frac{24}{2}$$

Divide -24 by 2.

$$x=-12$$

The equation is now solved.

$$x=10 x=-12$$

Steps for Completing the Square

Divide both sides by 6.

$$\left(1+x\right)^{2}=\frac{726}{6}$$

Divide 726 by 6 to get 121.

$$\left(1+x\right)^{2}=121$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$1+2x+x^{2}=121$$

Subtract 1 from both sides.

$$2x+x^{2}=121-1$$

Subtract 1 from 121 to get 120.

$$2x+x^{2}=120$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+2x=120$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=120+1^{2}$$

Square 1.

$$x^{2}+2x+1=120+1$$

Add 120 to 1.

$$x^{2}+2x+1=121$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=121$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{121}$$

Simplify.

$$x+1=11 x+1=-11$$

Subtract 1 from both sides of the equation.

$$x=10 x=-12$$