Solve (2-w)(2-w^2)=7

$$( 2 - w ) ( 2 - w ^ { 2 } ) = 7$$

Solve for w

$$w=3$$

Solution Steps

Use the distributive property to multiply 2-w by 2-w^{2}.

$$4-2w^{2}-2w+w^{3}=7$$

Subtract 7 from both sides.

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

Subtract 7 from 4 to get -3.

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

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

$$w^{3}-2w^{2}-2w-3=0$$

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

$$±3,±1$$

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.

$$w=3$$

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

$$w^{2}+w+1=0$$

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

$$w=\frac{-1±\sqrt{1^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$w=\frac{-1±\sqrt{-3}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$w\in \emptyset$$

List all found solutions.

$$w=3$$