Solve a^2-4a-4a+16

$$a ^ { 2 } - 4 a - 4 a + 16$$

Factor

$$\left(a-4\right)^{2}$$

Solution Steps

Multiply and combine like terms.

$$a^{2}-8a+16$$

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+16. To find p and q, set up a system to be solved.

$$p+q=-8 pq=1\times 16=16$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$p=-4 q=-4$$

Rewrite a^{2}-8a+16 as \left(a^{2}-4a\right)+\left(-4a+16\right).

$$\left(a^{2}-4a\right)+\left(-4a+16\right)$$

Factor out a in the first and -4 in the second group.

$$a\left(a-4\right)-4\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(a-4\right)$$

Rewrite as a binomial square.

$$\left(a-4\right)^{2}$$

Evaluate

$$\left(a-4\right)^{2}$$

Solution Steps

Combine -4a and -4a to get -8a.

$$a^{2}-8a+16$$