Polynomials!!!!

Question

Find all zeroes of the function and wrote the polynomial as a product of linear factors.

f(x)=x^3-x^2+36x-36

Answer 1

Sometimes you can find the zeroes by just trying the simplest integers first.

Let x=0, that doesn't work, as the polynomial is -36

Try x=1, replace x by 1. That works 1³-1²+36(1)-36 = 1-1+36-36=0

x-1 is a factor of the 3rd degree given polynomial. x=1 is a zero

Divide that into the 3 degree polynomial to get x²+36. You can use long division

(x-1)(x²+36) = x³-x²+36x-36

x²+36=0 or x²=-36 means x= plus or minus an imaginary number -(-36)^1/2 or (-36)^1/2

That's equal to -6i or +6i

There is only one zero x=1. If you graphed the polynomial it would touch or intersect the x axis in only one place, the point (1,0)

Answer 2

With any polynomial 3rd degree or higher, we start with long division or synthetic division to find the first terms. This is a trial and error process.

In this case, it is relatively easy to see that x - 1 is a factor. Divide the expression by x - 1, and we have:

x^2 + 36, which in turn has 2 imaginary factors