A line that includes the points (9,j) and (10,4) has a slope of 1. What is the value of j?

j = 3

Step-by-step explanation:

(9 , j) ;   and

(10, 4) ;   and

4 - j = 1   {Subtract 4 from both sides}

-j = 1 -4

-j = -3 {Multiply both sides by (-1) }

j = 3

f(x) = 4x + 11x2 - 104x + 105  =  (x - 3)(x + 7)(4x - 5)

Step-by-step explanation:

f(x) = 4x^3 + 11x2 - 104x + 105

has the coefficients {4, 11 -104, 105}.  We perform synthetic division using the given -7 as divisor:

-7   /   4      11     -104     105

                -28     119     -105

          4      -17      15         0

Since the remainder is zero (0), we have shown that -7 is a zero of f(x).  The coefficients of the quotient (above) are {4, -17, 15}.  Let's try factoring the corresponding polynomial again using synthetic division.  Start out by using 5 as divisor and determining whether or not the remaindeer is zero:

5    /     4     -17      15

                    20     15

             4        3     30           No, the remainder is 30 and so 5 is not a

                                               root of this quadratic.  Try the divisor 3 instead:

3     /     4      -17      15

                      12      -15

             4       -5        0        Yes, the remainder is 0 and so 3 is a root.

Thus, the given f(x) = 4x + 11x2 - 104x + 105 factors as follows:

f(x) = 4x + 11x2 - 104x + 105  =  (x - 3)(x + 7)(4x - 5).  Notice that the coefficients of the last factor come from those we found above when using 3 as a divisor in synthetic division.