Solve =6x^3+13x^2-13x-20

$$= 6 x ^ { 3 } + 13 x ^ { 2 } - 13 x - 20$$

Factor

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

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 -20 and q divides the leading coefficient 6. One such root is -\frac{5}{2}. Factor the polynomial by dividing it by 2x+5.

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

Consider 3x^{2}-x-4. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-4\right)=-12$$

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 -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 -1.

$$a=-4 b=3$$

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

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

Factor out x in 3x^{2}-4x.

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

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x+1\right)$$

Rewrite the complete factored expression.

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

Evaluate

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