Notice that the first two terms of the expression have x² as a common factor, and the last two terms have 36 as a common factor.
Factor out x² from the first two terms, and -36 from the last two terms:
[tex]\begin{gathered} x^3-x^2-36x+36=(x^3-x^2)+(-36x+36) \\ =x^2(x-1)-36(x-1) \end{gathered}[/tex]Now, we are left with two terms that have (x-1) as a common factor. Factor out (x-1) from the expression:
[tex]x^2(x-1)-36(x-1)=(x-1)(x^2-36)[/tex]Notice that 36 is equal to 6². Rewrite the expression using this fact:
[tex](x-1)(x^2-36)=(x-1)(x^2-6^2)[/tex]The second factor of the expression is a difference of squares. Remember that a difference of squares can be factored out as the product of two conjugate binomials, as follows:
[tex](a^2-b^2)=(a+b)(a-b)[/tex]Then, factor out the difference of squares x²-6² as a product of two conjugate binomials. Be careful to place the minus sign correctly:
[tex](x-1)(x^2-6^2)=(x-1)(x+6)(x-6)[/tex]Then, the complete factorization of the polynomial is:
[tex]x^3-x^2-36x+36=(x-1)(x+6)(x-6)[/tex]To solve the equation, remember that any expression multiplied by 0 is equal to 0. So, if the product of three factors is equal to 0, there are three possibilities, since the whole expression will be equal to 0 whenever any of the factors is equal to 0:
[tex]\begin{gathered} x^3-x^2-36x+36=0 \\ \Rightarrow(x-1)(x+6)(x-6)=0 \\ \Rightarrow x-1=0 \\ OR\colon x+6=0 \\ OR\colon x-6=0 \\ \Rightarrow x=1 \\ OR\colon x=-6 \\ OR\colon x=6 \end{gathered}[/tex]Therefore, the solution set for this equation is {-6,1,6}. The complete factorization of the given polinomial is (x-1)(x+6)(x-6).
