Solve 4yz^3+3yz^2-10yz

$$4 y z ^ { 3 } + 3 y z ^ { 2 } - 10 y z$$

Factor

$$yz\left(4z-5\right)\left(z+2\right)$$

Solution Steps

Factor out yz.

$$yz\left(4z^{2}+3z-10\right)$$

Consider 4z^{2}+3z-10. Factor the expression by grouping. First, the expression needs to be rewritten as 4z^{2}+az+bz-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=4\left(-10\right)=-40$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=-5 b=8$$

Rewrite 4z^{2}+3z-10 as \left(4z^{2}-5z\right)+\left(8z-10\right).

$$\left(4z^{2}-5z\right)+\left(8z-10\right)$$

Factor out z in the first and 2 in the second group.

$$z\left(4z-5\right)+2\left(4z-5\right)$$

Factor out common term 4z-5 by using distributive property.

$$\left(4z-5\right)\left(z+2\right)$$

Rewrite the complete factored expression.

$$yz\left(4z-5\right)\left(z+2\right)$$

Evaluate

$$yz\left(4z-5\right)\left(z+2\right)$$