Solve 3y^3-7y^2+4=0

$$3 y ^ { 3 } - 7 y ^ { 2 } + 4 = 0$$

Solve for y

$$y=1$$

$$y=2$$

$$y=-\frac{2}{3}\approx -0.666666667$$

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 4 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$y=1$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 3y^{3}-7y^{2}+4 by y-1 to get 3y^{2}-4y-4. Solve the equation where the result equals to 0.

$$3y^{2}-4y-4=0$$

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 3 for a, -4 for b, and -4 for c in the quadratic formula.

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

Do the calculations.

$$y=\frac{4±8}{6}$$

Solve the equation 3y^{2}-4y-4=0 when ± is plus and when ± is minus.

$$y=-\frac{2}{3} y=2$$

List all found solutions.

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