Solve 2x^4+7x^3-26x^2+23x-6

$$2 { x }^{ 4 } +7 { x }^{ 3 } -26 { x }^{ 2 } +23x-6$$

Factor

$$\left(2x-1\right)\left(x+6\right)\left(x-1\right)^{2}$$

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 -6 and q divides the leading coefficient 2. One such root is 1. Factor the polynomial by dividing it by x-1.

$$\left(x-1\right)\left(2x^{3}+9x^{2}-17x+6\right)$$

Consider 2x^{3}+9x^{2}-17x+6. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 2. One such root is 1. Factor the polynomial by dividing it by x-1.

$$\left(x-1\right)\left(2x^{2}+11x-6\right)$$

Consider 2x^{2}+11x-6. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=11 ab=2\left(-6\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 11.

$$a=-1 b=12$$

Rewrite 2x^{2}+11x-6 as \left(2x^{2}-x\right)+\left(12x-6\right).

$$\left(2x^{2}-x\right)+\left(12x-6\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(2x-1\right)+6\left(2x-1\right)$$

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

$$\left(2x-1\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$\left(2x-1\right)\left(x+6\right)\left(x-1\right)^{2}$$

Evaluate

$$\left(2x-1\right)\left(x+6\right)\left(x-1\right)^{2}$$