Solve x(3x^2)+x(x)-x(2)=

$$x ( 3 x ^ { 2 } ) + x ( x ) - x ( 2 ) =$$

Evaluate

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

Solution Steps

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$x^{3}\times 3+xx-x\times 2$$

Multiply x and x to get x^{2}.

$$x^{3}\times 3+x^{2}-x\times 2$$

Multiply -1 and 2 to get -2.

$$x^{3}\times 3+x^{2}-2x$$

Factor

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

Solution Steps

Factor out x.

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

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

$$a+b=1 ab=3\left(-2\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

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

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

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

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

Factor out common term 3x-2 by using distributive property.

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

Rewrite the complete factored expression.

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