Solve 2t^5+4t^4-10t^3-12t^2

$$2 t ^ { 5 } + 4 t ^ { 4 } - 10 t ^ { 3 } - 12 t ^ { 2 }$$

Factor

$$2\left(t-2\right)\left(t+1\right)\left(t+3\right)t^{2}$$

Solution Steps

Factor out 2.

$$2\left(t^{5}+2t^{4}-5t^{3}-6t^{2}\right)$$

Consider t^{5}+2t^{4}-5t^{3}-6t^{2}. Factor out t^{2}.

$$t^{2}\left(t^{3}+2t^{2}-5t-6\right)$$

Consider t^{3}+2t^{2}-5t-6. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. One such root is -3. Factor the polynomial by dividing it by t+3.

$$\left(t+3\right)\left(t^{2}-t-2\right)$$

Consider t^{2}-t-2. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-2. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite t^{2}-t-2 as \left(t^{2}-2t\right)+\left(t-2\right).

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

Factor out t in t^{2}-2t.

$$t\left(t-2\right)+t-2$$

Factor out common term t-2 by using distributive property.

$$\left(t-2\right)\left(t+1\right)$$

Rewrite the complete factored expression.

$$2t^{2}\left(t+3\right)\left(t-2\right)\left(t+1\right)$$

Evaluate

$$2\left(t-2\right)\left(t+1\right)\left(t+3\right)t^{2}$$