Solve y^3-7y^2+16y-12=0

$$y ^ { 3 } - 7 y ^ { 2 } + 16 y - 12 = 0$$

Solve for y

$$y=2$$

$$y=3$$

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 -12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$y=2$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}-7y^{2}+16y-12 by y-2 to get y^{2}-5y+6. Solve the equation where the result equals to 0.

$$y^{2}-5y+6=0$$

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 1 for a, -5 for b, and 6 for c in the quadratic formula.

$$y=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$y=\frac{5±1}{2}$$

Solve the equation y^{2}-5y+6=0 when ± is plus and when ± is minus.

$$y=2 y=3$$

List all found solutions.

$$y=2 y=3$$