Solve 2x^3+3x^2-6x-4div2x+3

$$2 x ^ { 3 } + 3 x ^ { 2 } - 6 x - 4 \div 2 x + 3$$

Evaluate

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

Solution Steps

Divide 4 by 2 to get 2.

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

Combine -6x and -2x to get -8x.

$$2x^{3}+3x^{2}-8x+3$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$2x^{3}+3x^{2}-8x+3$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 3 and q divides the leading coefficient 2. One such root is \frac{1}{2}. Factor the polynomial by dividing it by 2x-1.

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

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

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

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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