Due in an hour Factor​ f(x) into linear factors given that k is a zero of​ f(x). F(x)=4×3+19×2-138x+135; k=3 Factor f(x) = 2×3+x2-71x+140 into linear factors given that -7 is a zero of f(x) For the polynomial function, (a) list all possible rational zeros, (b) find all rational zeros, and (c) factor f(x) F(x) = x3+3×2-25x+21 Purchase the answer to view it

To factor the polynomial function (f(x) = 4x^3 + 19x^2 – 138x + 135) using a given zero (k = 3), we can use synthetic division or long division to divide (f(x)) by ((x – k)).

Using synthetic division:

3 | 4 19 -138 135
| 12 93 -135
———————-
4 31 -45 0

The result of the synthetic division is (4x^2 + 31x – 45). Therefore, we can factor (f(x)) as

[f(x) = (x – 3)(4x^2 + 31x – 45).]

To factor the second polynomial function (f(x) = 2x^3 + x^2 – 71x + 140) using the zero (-7), we follow the same process:

-7 | 2 1 -71 140
| -14 91 -20
——————
2 -13 20 120

The result of the synthetic division is (2x^2 – 13x + 20). Therefore, we can factor (f(x)) as

[f(x) = (x + 7)(2x^2 – 13x + 20).]

Now, for the polynomial function (f(x) = x^3 + 3x^2 – 25x + 21), we are asked to (a) list all possible rational zeros, (b) find all the rational zeros, and (c) factor (f(x)).

(a) To list all possible rational zeros, we can use the Rational Root Theorem. According to the theorem, the possible rational zeros of a polynomial function are in the form (frac{p}{q}), where (p) is a factor of the constant term and (q) is a factor of the leading coefficient.

In this case, the constant term is 21 and the leading coefficient is 1. The factors of 21 are (pm 1, pm 3, pm 7, pm 21), and the factors of 1 are (pm 1). Therefore, the possible rational zeros are (pm 1, pm 3, pm 7, pm 21).

(b) To find the rational zeros, we can use synthetic division or long division to test each possible zero. Starting with the first possible zero (-1):

-1 | 1 3 -25 21
| -1 -2 27
———————
1 2 -27 48

The result of the synthetic division is (x^2 + 2x – 27). The next possible zero is (-3):

-3 | 1 2 -27 48
| -3 3 72
———————-
1 -1 -24 120

The result of the synthetic division is (x^2 – x – 24). The next possible zero is (-7):

-7 | 1 -1 -24 120
| -7 56 -16
———————
1 -8 32 104

The result of the synthetic division is (x^2 – 8x + 32). None of the possible zeros (pm 1, pm 3, pm 7, pm 21) are actual zeros of the function because the remainders in each synthetic division are not zero. This means that the polynomial function (f(x) = x^3 + 3x^2 – 25x + 21) does not have any rational zeros.

(c) To factor (f(x)), we can use the polynomial obtained from synthetic division, which is (x^2 – 8x + 32). However, this polynomial cannot be factored further using rational numbers. Therefore, the factored form of (f(x)) is

[f(x) = (x^2 – 8x + 32)(text{irreducible quadratic}).]

In conclusion, the factored forms of the given polynomial functions are:

1. (f(x) = 4x^3 + 19x^2 – 138x + 135) given that (k = 3):
[f(x) = (x – 3)(4x^2 + 31x – 45).]

2. (f(x) = 2x^3 + x^2 – 71x + 140) given that (k = -7):
[f(x) = (x + 7)(2x^2 – 13x + 20).]

3. (f(x) = x^3 + 3x^2 – 25x + 21):
[f(x)) does not have any rational zeros and cannot be factored further using rational numbers.