Solve (x+1)^2-5x(x+1)=0

$$( x + 1 ) ^ { 2 } - 5 x ( x + 1 ) = 0$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=-1$$

Steps Using Factoring By Grouping

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

$$x^{2}+2x+1-5x\left(x+1\right)=0$$

Use the distributive property to multiply -5x by x+1.

$$x^{2}+2x+1-5x^{2}-5x=0$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$-4x^{2}+2x+1-5x=0$$

Combine 2x and -5x to get -3x.

$$-4x^{2}-3x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-4=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=1 b=-4$$

Rewrite -4x^{2}-3x+1 as \left(-4x^{2}+x\right)+\left(-4x+1\right).

$$\left(-4x^{2}+x\right)+\left(-4x+1\right)$$

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

$$-x\left(4x-1\right)-\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(-x-1\right)$$

To find equation solutions, solve 4x-1=0 and -x-1=0.

$$x=\frac{1}{4} x=-1$$

Steps Using the Quadratic Formula

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

$$x^{2}+2x+1-5x\left(x+1\right)=0$$

Use the distributive property to multiply -5x by x+1.

$$x^{2}+2x+1-5x^{2}-5x=0$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$-4x^{2}+2x+1-5x=0$$

Combine 2x and -5x to get -3x.

$$-4x^{2}-3x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-4\right)}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-4\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\left(-4\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{3±5}{-8}$$

Now solve the equation x=\frac{3±5}{-8} when ± is plus. Add 3 to 5.

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

Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{3±5}{-8} when ± is minus. Subtract 5 from 3.

$$x=-\frac{2}{-8}$$

Reduce the fraction \frac{-2}{-8} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{4}$$

The equation is now solved.

$$x=-1 x=\frac{1}{4}$$

Steps for Completing the Square

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

$$x^{2}+2x+1-5x\left(x+1\right)=0$$

Use the distributive property to multiply -5x by x+1.

$$x^{2}+2x+1-5x^{2}-5x=0$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$-4x^{2}+2x+1-5x=0$$

Combine 2x and -5x to get -3x.

$$-4x^{2}-3x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}-3x=-1$$

Divide both sides by -4.

$$\frac{-4x^{2}-3x}{-4}=-\frac{1}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{3}{-4}\right)x=-\frac{1}{-4}$$

Divide -3 by -4.

$$x^{2}+\frac{3}{4}x=-\frac{1}{-4}$$

Divide -1 by -4.

$$x^{2}+\frac{3}{4}x=\frac{1}{4}$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\frac{1}{4}+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{1}{4}+\frac{9}{64}$$

Add \frac{1}{4} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{25}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{5}{8} x+\frac{3}{8}=-\frac{5}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{1}{4} x=-1$$