Solve 4x^2-2*2x*5+5^2

$$4 x ^ { 2 } - 2 \times 2 x \times 5 + 5 ^ { 2 }$$

Evaluate

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

Solution Steps

Multiply 2 and 2 to get 4.

$$4x^{2}-4x\times 5+5^{2}$$

Multiply 4 and 5 to get 20.

$$4x^{2}-20x+5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$4x^{2}-20x+25$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$4x^{2}-20x+25$$

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-20 ab=4\times 25=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -20.

$$a=-10 b=-10$$

Rewrite 4x^{2}-20x+25 as \left(4x^{2}-10x\right)+\left(-10x+25\right).

$$\left(4x^{2}-10x\right)+\left(-10x+25\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(2x-5\right)-5\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(2x-5\right)$$

Rewrite as a binomial square.

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