Solve g(x)leq0=2x^3-x^2-4x+3

$$g ( x ) \leq 0 = 2 x ^ { 3 } - x ^ { 2 } - 4 x + 3$$

Solve for x

$$x=1$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

$$2x^{3}-x^{2}-4x+3=0$$

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. List all candidates \frac{p}{q}.

$$±\frac{3}{2},±3,±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-x^{2}-4x+3 by x-1 to get 2x^{2}+x-3. Solve the equation where the result equals to 0.

$$2x^{2}+x-3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, 1 for b, and -3 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-1±5}{4}$$

Solve the equation 2x^{2}+x-3=0 when ± is plus and when ± is minus.

$$x=-\frac{3}{2} x=1$$

List all found solutions.

$$x=1 x=-\frac{3}{2}$$