Solve 2x^2z+10xz+8z

$$2 x ^ { 2 } z + 10 x z + 8 z$$

Factor

$$2z\left(x+1\right)\left(x+4\right)$$

Solution Steps

Factor out 2.

$$2\left(x^{2}z+5xz+4z\right)$$

Consider x^{2}z+5xz+4z. Factor out z.

$$z\left(x^{2}+5x+4\right)$$

Consider x^{2}+5x+4. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=1 b=4$$

Rewrite x^{2}+5x+4 as \left(x^{2}+x\right)+\left(4x+4\right).

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

Factor out x in the first and 4 in the second group.

$$x\left(x+1\right)+4\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$2z\left(x+1\right)\left(x+4\right)$$

Evaluate

$$2z\left(x+1\right)\left(x+4\right)$$