Solve (x-4)^2*(x+3)^3*(x-1)=0

$$( x - 4 ) ^ { 2 } \cdot ( x + 3 ) ^ { 3 } \cdot ( x - 1 ) = 0$$

Solve for x

$$x=-3$$

$$x=4$$

$$x=1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$\left(x^{2}-8x+16\right)\left(x+3\right)^{3}\left(x-1\right)=0$$

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

$$\left(x^{2}-8x+16\right)\left(x^{3}+9x^{2}+27x+27\right)\left(x-1\right)=0$$

Use the distributive property to multiply x^{2}-8x+16 by x^{3}+9x^{2}+27x+27 and combine like terms.

$$\left(x^{5}+x^{4}-29x^{3}-45x^{2}+216x+432\right)\left(x-1\right)=0$$

Use the distributive property to multiply x^{5}+x^{4}-29x^{3}-45x^{2}+216x+432 by x-1 and combine like terms.

$$x^{6}-30x^{4}-16x^{3}+261x^{2}+216x-432=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -432 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±432,±216,±144,±108,±72,±54,±48,±36,±27,±24,±18,±16,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{6}-30x^{4}-16x^{3}+261x^{2}+216x-432 by x-1 to get x^{5}+x^{4}-29x^{3}-45x^{2}+216x+432. Solve the equation where the result equals to 0.

$$x^{5}+x^{4}-29x^{3}-45x^{2}+216x+432=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 432 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±432,±216,±144,±108,±72,±54,±48,±36,±27,±24,±18,±16,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{5}+x^{4}-29x^{3}-45x^{2}+216x+432 by x+3 to get x^{4}-2x^{3}-23x^{2}+24x+144. Solve the equation where the result equals to 0.

$$x^{4}-2x^{3}-23x^{2}+24x+144=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 144 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±144,±72,±48,±36,±24,±18,±16,±12,±9,±8,±6,±4,±3,±2,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-2x^{3}-23x^{2}+24x+144 by x+3 to get x^{3}-5x^{2}-8x+48. Solve the equation where the result equals to 0.

$$x^{3}-5x^{2}-8x+48=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 48 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±48,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-5x^{2}-8x+48 by x+3 to get x^{2}-8x+16. Solve the equation where the result equals to 0.

$$x^{2}-8x+16=0$$

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, -8 for b, and 16 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{8±0}{2}$$

Solutions are the same.

$$x=4$$

List all found solutions.

$$x=1 x=-3 x=4$$