Solve g(x)=6x^3-41x^2-8x+7

$$g ( x ) = 6 x ^ { 3 } - 41 x ^ { 2 } - 8 x + 7$$

Factor

$$\left(x-7\right)\left(3x-1\right)\left(2x+1\right)$$

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

$$\left(2x+1\right)\left(3x^{2}-22x+7\right)$$

Consider 3x^{2}-22x+7. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=-22 ab=3\times 7=21$$

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 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -22.

$$a=-21 b=-1$$

Rewrite 3x^{2}-22x+7 as \left(3x^{2}-21x\right)+\left(-x+7\right).

$$\left(3x^{2}-21x\right)+\left(-x+7\right)$$

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

$$3x\left(x-7\right)-\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(3x-1\right)$$

Rewrite the complete factored expression.

$$\left(x-7\right)\left(3x-1\right)\left(2x+1\right)$$

Evaluate

$$\left(x-7\right)\left(3x-1\right)\left(2x+1\right)$$