Solve -4x^4+16x^3+128x^2

$$- 4 x ^ { 4 } + 16 x ^ { 3 } + 128 x ^ { 2 }$$

Factor

$$-4\left(x-8\right)\left(x+4\right)x^{2}$$

Solution Steps

Factor out 4.

$$4\left(-x^{4}+4x^{3}+32x^{2}\right)$$

Consider -x^{4}+4x^{3}+32x^{2}. Factor out x^{2}.

$$x^{2}\left(-x^{2}+4x+32\right)$$

Consider -x^{2}+4x+32. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+32. To find a and b, set up a system to be solved.

$$a+b=4 ab=-32=-32$$

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 -32.

$$-1,32 -2,16 -4,8$$

Calculate the sum for each pair.

$$-1+32=31 -2+16=14 -4+8=4$$

The solution is the pair that gives sum 4.

$$a=8 b=-4$$

Rewrite -x^{2}+4x+32 as \left(-x^{2}+8x\right)+\left(-4x+32\right).

$$\left(-x^{2}+8x\right)+\left(-4x+32\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(x-8\right)-4\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x-4\right)$$

Rewrite the complete factored expression.

$$4x^{2}\left(x-8\right)\left(-x-4\right)$$

Evaluate

$$-4\left(x-8\right)\left(x+4\right)x^{2}$$