Solve x^2-2*x*5+5^2

$$x ^ { 2 } - 2 \cdot x \cdot 5 + 5 ^ { 2 }$$

Evaluate

$$\left(x-5\right)^{2}$$

Solution Steps

Multiply 2 and 5 to get 10.

$$x^{2}-10x+5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$x^{2}-10x+25$$

Factor

$$\left(x-5\right)^{2}$$

Solution Steps

Multiply and combine like terms.

$$x^{2}-10x+25$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 25=25$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -10.

$$a=-5 b=-5$$

Rewrite x^{2}-10x+25 as \left(x^{2}-5x\right)+\left(-5x+25\right).

$$\left(x^{2}-5x\right)+\left(-5x+25\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-5\right)-5\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-5\right)$$

Rewrite as a binomial square.

$$\left(x-5\right)^{2}$$