Solve 8x^4+6x^3-19x^2+3x+2

$$8 { x }^{ 4 } +6 { x }^{ 3 } -19 { x }^{ 2 } +3x+2$$

Factor

$$\left(x-1\right)\left(2x-1\right)\left(x+2\right)\left(4x+1\right)$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 8. One such root is -\frac{1}{4}. Factor the polynomial by dividing it by 4x+1.

$$\left(4x+1\right)\left(2x^{3}+x^{2}-5x+2\right)$$

Consider 2x^{3}+x^{2}-5x+2. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 2. One such root is 1. Factor the polynomial by dividing it by x-1.

$$\left(x-1\right)\left(2x^{2}+3x-2\right)$$

Consider 2x^{2}+3x-2. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=3 ab=2\left(-2\right)=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite 2x^{2}+3x-2 as \left(2x^{2}-x\right)+\left(4x-2\right).

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

Factor out x in the first and 2 in the second group.

$$x\left(2x-1\right)+2\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$\left(x-1\right)\left(2x-1\right)\left(4x+1\right)\left(x+2\right)$$

Evaluate

$$\left(x-1\right)\left(2x-1\right)\left(x+2\right)\left(4x+1\right)$$