Solve 2z^4+12z^3-14z^2

$$2 z ^ { 4 } + 12 z ^ { 3 } - 14 z ^ { 2 }$$

Factor

$$2\left(z-1\right)\left(z+7\right)z^{2}$$

Solution Steps

Factor out 2.

$$2\left(z^{4}+6z^{3}-7z^{2}\right)$$

Consider z^{4}+6z^{3}-7z^{2}. Factor out z^{2}.

$$z^{2}\left(z^{2}+6z-7\right)$$

Consider z^{2}+6z-7. Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-7. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-7\right)=-7$$

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. The only such pair is the system solution.

$$a=-1 b=7$$

Rewrite z^{2}+6z-7 as \left(z^{2}-z\right)+\left(7z-7\right).

$$\left(z^{2}-z\right)+\left(7z-7\right)$$

Factor out z in the first and 7 in the second group.

$$z\left(z-1\right)+7\left(z-1\right)$$

Factor out common term z-1 by using distributive property.

$$\left(z-1\right)\left(z+7\right)$$

Rewrite the complete factored expression.

$$2z^{2}\left(z-1\right)\left(z+7\right)$$

Evaluate

$$2\left(z-1\right)\left(z+7\right)z^{2}$$