Solve 4left(n-1right)^2=0

$$4 { \left(n-1 \right) }^{ 2 } =0$$

Solve for n

$$n=1$$

Steps Using Factoring

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$\left(n-1\right)^{2}=0$$

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

$$n^{2}-2n+1=0$$

To solve the equation, factor n^{2}-2n+1 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=1$$

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.

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

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

$$\left(n-1\right)\left(n-1\right)$$

Rewrite as a binomial square.

$$\left(n-1\right)^{2}$$

To find equation solution, solve n-1=0.

$$n=1$$

Steps Using Factoring By Grouping

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$\left(n-1\right)^{2}=0$$

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

$$n^{2}-2n+1=0$$

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

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

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.

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

Rewrite n^{2}-2n+1 as \left(n^{2}-n\right)+\left(-n+1\right).

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

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

$$n\left(n-1\right)-\left(n-1\right)$$

Factor out common term n-1 by using distributive property.

$$\left(n-1\right)\left(n-1\right)$$

Rewrite as a binomial square.

$$\left(n-1\right)^{2}$$

To find equation solution, solve n-1=0.

$$n=1$$

Steps Using the Quadratic Formula

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$\left(n-1\right)^{2}=0$$

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

$$n^{2}-2n+1=0$$

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

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

Square -2.

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

Add 4 to -4.

$$n=\frac{-\left(-2\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$n=-\frac{-2}{2}$$

The opposite of -2 is 2.

$$n=\frac{2}{2}$$

Divide 2 by 2.

$$n=1$$

Steps for Completing the Square

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$\left(n-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$n-1=0 n-1=0$$

Add 1 to both sides of the equation.

$$n=1 n=1$$

The equation is now solved. Solutions are the same.

$$n=1$$