Solve 4y^3-7y^2+28-16y=0

$$4 y ^ { 3 } - 7 y ^ { 2 } + 28 - 16 y = 0$$

Solve for y

$$y=2$$

$$y = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

$$y=-2$$

Solution Steps

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$4y^{3}-7y^{2}-16y+28=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 28 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±7,±14,±28,±\frac{7}{2},±\frac{7}{4},±1,±2,±4,±\frac{1}{2},±\frac{1}{4}$$

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 4y^{3}-7y^{2}-16y+28 by y-2 to get 4y^{2}+y-14. Solve the equation where the result equals to 0.

$$4y^{2}+y-14=0$$

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 4 for a, 1 for b, and -14 for c in the quadratic formula.

$$y=\frac{-1±\sqrt{1^{2}-4\times 4\left(-14\right)}}{2\times 4}$$

Do the calculations.

$$y=\frac{-1±15}{8}$$

Solve the equation 4y^{2}+y-14=0 when ± is plus and when ± is minus.

$$y=-2 y=\frac{7}{4}$$

List all found solutions.

$$y=2 y=-2 y=\frac{7}{4}$$