Solve 48x^5+20x^4-100x^3

$$48 x ^ { 5 } + 20 x ^ { 4 } - 100 x ^ { 3 }$$

Factor

$$4\left(4x-5\right)\left(3x+5\right)x^{3}$$

Solution Steps

Factor out 4.

$$4\left(12x^{5}+5x^{4}-25x^{3}\right)$$

Consider 12x^{5}+5x^{4}-25x^{3}. Factor out x^{3}.

$$x^{3}\left(12x^{2}+5x-25\right)$$

Consider 12x^{2}+5x-25. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-25. To find a and b, set up a system to be solved.

$$a+b=5 ab=12\left(-25\right)=-300$$

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 -300.

$$-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20$$

Calculate the sum for each pair.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 5.

$$a=-15 b=20$$

Rewrite 12x^{2}+5x-25 as \left(12x^{2}-15x\right)+\left(20x-25\right).

$$\left(12x^{2}-15x\right)+\left(20x-25\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(4x-5\right)+5\left(4x-5\right)$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(3x+5\right)$$

Rewrite the complete factored expression.

$$4x^{3}\left(4x-5\right)\left(3x+5\right)$$

Evaluate

$$4\left(4x-5\right)\left(3x+5\right)x^{3}$$