Question

find all real zeros of this polynomial
x^3-2x^2-9x+18

Answer 1

First, try to factor completely.
Is is best to separate in two parts:

x³-2x²-9x+18 = (x³-2x²)-(9x-18) = x²(x-2) -9(x-2).

Can you see now how to factor out one more common factor?

Answer 2

so you can group these together
(x^3 - 3x^2)(-9x+18)
what can come out of each?

x^2(x - 3) -9(x -2)
can you keep going?

Answer 3

an 8 can come out of each

Answer 4

x*

Answer 5

and sorry for my typo inthe first part

should be

(x^3 - 2x^2)(-9x+18)

becomes

x^2(x - 2) -9(x -2)

Answer 6

Once you have x²(x-2) -9(x-2), you can see that x-2 is common, so it can be written as:

(x-2)(x²-9).

But: x²-9 can itself be further factored. It is of the form a²-b²=(a+b)(a-b),
so it is (x+3)(x-3)

Putting everything together, we have: (x-2)(x+3)(x-3).

To get the zeroes, solve: (x-2)(x+3)(x-3)=0.
This is easy now...

Answer 7

okay that makes sense

Answer 8

So there are your three real zeroes!

Answer 9

thanks so much

Answer 10

YW!