Solve 6x^3+7x^2-3x

$$6 x ^ { 3 } + 7 x ^ { 2 } - 3 x$$

Factor

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

Solution Steps

Factor out x.

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

Consider 6x^{2}+7x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

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

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 -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.

$$a=-2 b=9$$

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

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

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

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

Factor out common term 3x-1 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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