Solve -2x^4+6x^3+20x^2

$$- 2 x ^ { 4 } + 6 x ^ { 3 } + 20 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out 2.

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

Consider -x^{4}+3x^{3}+10x^{2}. Factor out x^{2}.

$$x^{2}\left(-x^{2}+3x+10\right)$$

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

$$a+b=3 ab=-10=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=5 b=-2$$

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

$$\left(-x^{2}+5x\right)+\left(-2x+10\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-5\right)-2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x-2\right)$$

Rewrite the complete factored expression.

$$2x^{2}\left(x-5\right)\left(-x-2\right)$$

Evaluate

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