Question

When the polynomial P(x) = 5x3 − 51x2 + kx − 9 is divided by x − 9, the remainder is 0. Which of the following is also a factor of P(x)?
A- x − 5
B- x + 1
C- x − 1
D- 5x + 1

Answer 1

Answer:

answer C:  (x - 1) is also a factor of P(x).

Step-by-step 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)