Solve =6x^3-17x^2-5x+6

$$= 6 x ^ { 3 } - 17 x ^ { 2 } - 5 x + 6$$

Factor

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

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

$$\left(2x-1\right)\left(3x^{2}-7x-6\right)$$

Consider 3x^{2}-7x-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.

$$a+b=-7 ab=3\left(-6\right)=-18$$

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 -18.

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

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -7.

$$a=-9 b=2$$

Rewrite 3x^{2}-7x-6 as \left(3x^{2}-9x\right)+\left(2x-6\right).

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

Factor out 3x in the first and 2 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

$$6x^{3}-17x^{2}-5x+6$$