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

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

Solve for x

$$x=1$$

$$x=-1$$

$$x=-3$$

Solution Steps

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

$$\left(x^{2}-1\right)\left(x^{2}+6x+9\right)=0$$

Use the distributive property to multiply x^{2}-1 by x^{2}+6x+9 and combine like terms.

$$x^{4}+6x^{3}+8x^{2}-6x-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}+8x^{2}-6x-9 by x-1 to get x^{3}+7x^{2}+15x+9. Solve the equation where the result equals to 0.

$$x^{3}+7x^{2}+15x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+7x^{2}+15x+9 by x+1 to get x^{2}+6x+9. Solve the equation where the result equals to 0.

$$x^{2}+6x+9=0$$

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, 6 for b, and 9 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-6±0}{2}$$

Solutions are the same.

$$x=-3$$

List all found solutions.

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