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

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

Factor

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

Solution Steps

Factor out x^{2}.

$$x^{2}\left(x^{2}+x-20\right)$$

Consider x^{2}+x-20. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-20\right)=-20$$

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 -20.

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

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite x^{2}+x-20 as \left(x^{2}-4x\right)+\left(5x-20\right).

$$\left(x^{2}-4x\right)+\left(5x-20\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-4\right)+5\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+5\right)$$

Rewrite the complete factored expression.

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

Evaluate

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