Answer:
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.
