Solve 4y^3-7=y-28y^2

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

Solve for y

$$y=-\frac{1}{2}=-0.5$$

$$y=\frac{1}{2}=0.5$$

$$y=-7$$

Solution Steps

Subtract y from both sides.

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

Add 28y^{2} to both sides.

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

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -7 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

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

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.

$$y=-7$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 4y^{3}+28y^{2}-y-7 by y+7 to get 4y^{2}-1. Solve the equation where the result equals to 0.

$$4y^{2}-1=0$$

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, 0 for b, and -1 for c in the quadratic formula.

$$y=\frac{0±\sqrt{0^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Do the calculations.

$$y=\frac{0±4}{8}$$

Solve the equation 4y^{2}-1=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$y=-7 y=-\frac{1}{2} y=\frac{1}{2}$$