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

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

Solve for x

$$x=1$$

$$x=2$$

Steps Using Factoring

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

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

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

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

Add 1 and 1 to get 2.

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

To solve the equation, factor x^{2}-3x+2 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.

$$a+b=-3 ab=2$$

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.

$$a=-2 b=-1$$

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

$$\left(x-2\right)\left(x-1\right)$$

To find equation solutions, solve x-2=0 and x-1=0.

$$x=2 x=1$$

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}.

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

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

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

Add 1 and 1 to get 2.

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

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

$$a+b=-3 ab=1\times 2=2$$

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.

$$a=-2 b=-1$$

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

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

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

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

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

$$\left(x-2\right)\left(x-1\right)$$

To find equation solutions, solve x-2=0 and x-1=0.

$$x=2 x=1$$

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}.

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

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

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

Add 1 and 1 to get 2.

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

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

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

Square -3.

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

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2}$$

Add 9 to -8.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-3\right)±1}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±1}{2}$$

Now solve the equation x=\frac{3±1}{2} when ± is plus. Add 3 to 1.

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

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{3±1}{2} when ± is minus. Subtract 1 from 3.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=2 x=1$$

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}.

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

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

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

Add 1 and 1 to get 2.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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

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

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

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

Add -2 to \frac{9}{4}.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$