00:01
So problem number 23 gives us the polynomial x to the cubed minus 2x squared plus x minus 2, and they give us 0, x equals 2.
00:17
Now, if x is equal to 2, we know that one of the factors is going to have to be x minus 2, because when we set that equal to 0, we get x equals 2.
00:26
Now, in order to express this polynomial as a product of linear and irreducible quadratic factors, we are going to do long division.
00:37
So doing long division, having this factor outside, we can do x cubed minus 2 x squared plus x minus 2 divided by x minus 2.
00:52
So for long division, we need to find a number that we can multiply to x in order to get x cubed.
01:01
X squared times x is x cubed.
01:06
So now we can just multiply this out.
01:08
So x squared times x, we write x cubed down here...