Solve 20x^2-38x^3-30x^4

$$20 x ^ { 2 } - 38 x ^ { 3 } - 30 x ^ { 4 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(10x^{2}-19x^{3}-15x^{4}\right)$$

Consider 10x^{2}-19x^{3}-15x^{4}. Factor out x^{2}.

$$x^{2}\left(10-19x-15x^{2}\right)$$

Consider 10-19x-15x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-15x^{2}-19x+10$$

Factor the expression by grouping. First, the expression needs to be rewritten as -15x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-19 ab=-15\times 10=-150$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -19.

$$a=6 b=-25$$

Rewrite -15x^{2}-19x+10 as \left(-15x^{2}+6x\right)+\left(-25x+10\right).

$$\left(-15x^{2}+6x\right)+\left(-25x+10\right)$$

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

$$3x\left(-5x+2\right)+5\left(-5x+2\right)$$

Factor out common term -5x+2 by using distributive property.

$$\left(-5x+2\right)\left(3x+5\right)$$

Rewrite the complete factored expression.

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

Evaluate

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