Answer:
answer C: (x - 1) is also a factor of P(x).
Explanation:
Synthetic division is the best approach here. Given that one factor is x - 9, we know that 9 is the appropriate divisor in synthetic division:
9 ) 5 -51 k -9
45 -54 (9k - 486)
----------------------------------
5 -6 (k - 54) (-9 + 9k - 486)
and this remainder must = 0. Find k: -9 + 9k - 486 = 0, or
9k = 486 + 9 = 495
Then k = 495/9 = 55
Look at the last line of synthetic division, above:
5 -6 (k - 54) 0
Substituting 55 for k, we get:
5 -6 1
These are the coefficients of the quotient obtained by
dividing P(x) by (x - 9). They correspond to 5x^2 - 6x + 1.
We must factor this result.
Let's start with 5x + 1, and check whether this is a factor of 5x^2 - 6x + 1 or not. If 5x + 1 is a factor, then the related root is -1/5. Let's use -1/5 as the divisor in synthetic div.:
-1/5 ) 5 -6 1
-1 7/5
----------------------------
5 -7 12/5 Here the remainder is not zero, so -1/5 is
not a root and 5x + 1 is not a factor.
Now try x - 5. Is this a factor of 5x^2 - 6x + 1? Use 5 as divisor in synth. div.:
5 ) 5 -6 1
25 95
------------------------
5 19 96 Same conclusion: x - 5 is not a factor.
Try x = -1:
-1 ) 5 -6 1
-5 11
----------------------
5 -11 12. The remainder is not zero, so (x + 1) is not a factor.
Finally, try x = 1:
1 ) 5 -6 1
5 -1
--------------------
5 -1 0
Finally, we get a zero remainder, and thus we know that x - 1 is a factor of P(x)
Answer C is correct: x - 1 is a factor of P(x)