Question

I need to know how to factor the polynomial completely, setting each factor equal to zero. Showing my work

Answer 1

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:

`\begin{gathered} x^3-x^2-36x+36=(x^3-x^2)+(-36x+36) \\ =x^2(x-1)-36(x-1) \end{gathered}`

Now, we are left with two terms that have (x-1) as a common factor. Factor out (x-1) from the expression:

`x^2(x-1)-36(x-1)=(x-1)(x^2-36)`

Notice that 36 is equal to 6². Rewrite the expression using this fact:

`(x-1)(x^2-36)=(x-1)(x^2-6^2)`

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:

`(a^2-b^2)=(a+b)(a-b)`

Then, factor out the difference of squares x²-6² as a product of two conjugate binomials. Be careful to place the minus sign correctly:

`(x-1)(x^2-6^2)=(x-1)(x+6)(x-6)`

Then, the complete factorization of the polynomial is:

`x^3-x^2-36x+36=(x-1)(x+6)(x-6)`

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:

`\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}`

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).