Solve 3x^5-16x^4-13x^3+6x^2

$$3 x ^ { 5 } - 16 x ^ { 4 } - 13 x ^ { 3 } + 6 x ^ { 2 }$$

Factor

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

Solution Steps

Factor out x^{2}.

$$x^{2}\left(3x^{3}-16x^{2}-13x+6\right)$$

Consider 3x^{3}-16x^{2}-13x+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 3. One such root is -1. Factor the polynomial by dividing it by x+1.

$$\left(x+1\right)\left(3x^{2}-19x+6\right)$$

Consider 3x^{2}-19x+6. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-19 ab=3\times 6=18$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -19.

$$a=-18 b=-1$$

Rewrite 3x^{2}-19x+6 as \left(3x^{2}-18x\right)+\left(-x+6\right).

$$\left(3x^{2}-18x\right)+\left(-x+6\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-6\right)-\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(3x-1\right)$$

Rewrite the complete factored expression.

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

Evaluate

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