Solve left(2-xright)^3left(2-5xright)=0

$${ \left(2-x \right) }^{ 3 } \left( 2-5x \right) = 0$$

Solve for x

$$x=\frac{2}{5}=0.4$$

$$x=2$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(2-x\right)^{3}.

$$\left(8-12x+6x^{2}-x^{3}\right)\left(2-5x\right)=0$$

Use the distributive property to multiply 8-12x+6x^{2}-x^{3} by 2-5x and combine like terms.

$$16-64x+72x^{2}-32x^{3}+5x^{4}=0$$

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

$$5x^{4}-32x^{3}+72x^{2}-64x+16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 16 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

$$±\frac{16}{5},±16,±\frac{8}{5},±8,±\frac{4}{5},±4,±\frac{2}{5},±2,±\frac{1}{5},±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 5x^{4}-32x^{3}+72x^{2}-64x+16 by x-2 to get 5x^{3}-22x^{2}+28x-8. Solve the equation where the result equals to 0.

$$5x^{3}-22x^{2}+28x-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 5. List all candidates \frac{p}{q}.

$$±\frac{8}{5},±8,±\frac{4}{5},±4,±\frac{2}{5},±2,±\frac{1}{5},±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 5x^{3}-22x^{2}+28x-8 by x-2 to get 5x^{2}-12x+4. Solve the equation where the result equals to 0.

$$5x^{2}-12x+4=0$$

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 5 for a, -12 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\times 4}}{2\times 5}$$

Do the calculations.

$$x=\frac{12±8}{10}$$

Solve the equation 5x^{2}-12x+4=0 when ± is plus and when ± is minus.

$$x=\frac{2}{5} x=2$$

List all found solutions.

$$x=2 x=\frac{2}{5}$$