Solve 8x^4+56x^3-240x^2

$$8 x ^ { 4 } + 56 x ^ { 3 } - 240 x ^ { 2 }$$

Factor

$$8\left(x-3\right)\left(x+10\right)x^{2}$$

Solution Steps

Factor out 8.

$$8\left(x^{4}+7x^{3}-30x^{2}\right)$$

Consider x^{4}+7x^{3}-30x^{2}. Factor out x^{2}.

$$x^{2}\left(x^{2}+7x-30\right)$$

Consider x^{2}+7x-30. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\left(-30\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite x^{2}+7x-30 as \left(x^{2}-3x\right)+\left(10x-30\right).

$$\left(x^{2}-3x\right)+\left(10x-30\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x-3\right)+10\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+10\right)$$

Rewrite the complete factored expression.

$$8x^{2}\left(x-3\right)\left(x+10\right)$$

Evaluate

$$8\left(x-3\right)\left(x+10\right)x^{2}$$