Can the polynomial in the numerator of the expression x^2-5x+7/x-9 be factored to derive (x − 9) as a factor

Step-by-step explanation:

We can find out by using synthetic division and the fact that x - 9 Might be a 0. If x = 9 is a 0 of that polynomial, then we will get no remainder when we divide the polynomial by 9:

9 |   1     -5     7

First thing to do is bring down the 1 and multiply it by the 9 and put that product up under the -5:

9 |  1     -5     7

9

     1

Now add down the column to get:

9 |  1   -5    7

9

     1    4

Now multiply 4 by 9 and put that product up under the 7:

9 |   1    -5     7

9__36

      1     4

Now add to see what the remainder is:

9 |   1     -5     7

9_36

      1      4    R43

we have a remainder of 43. Since it is not a 0, then the polynomial is not evenly divisible by x - 9.

So the answer is No.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,