Use the rational zeros theorem to find all the real zeros of the polynomial function. Use the zeros to factor f over the real numbers.

f(x)= x^4+6x^3-11x^2-24x+28

Molon

well, 28 has to have factors

2 14
4 7

lets try 2 and -2
16 + 48 -44 - 48 +28
lo and behold = 0
so x = 2 is a solution and
(x-2) is a factor
so divide that mess by (x-2)
I get with long division
(to be continued :)

I started with x = ±1 and got x= 1

so (x-1) is a factor
and also found x = ±2 to work
so (x+2)(x-2) were factors
so

x^4+6x^3-11x^2-24x+28
= (x-1)(x-2)(x+2)(x .... ?)

so far my constant multiply to get 4, but I need +28, so the last digit must be +7

x^4+6x^3-11x^2-24x+28
= (x-1)(x-2)(x+2)(x + 7)

I get

(x-2)(x^3 + 8 x^2 + 5 x - 14)

now 14 is either 14 and 1 or 7 and 2
2 does not work this time :(
try -2
-8 + 32 -10 - 14 =-24+24 = 0
CARAMBA !!!
X = -2 IS A SOLUTION
SO
(x+2)(x-2) so far
divide x^3+8x^2+5x-14 by (x+2)

I GET x^2 + 6 x -7

facor that
(x+7)(x-1)
so -7 and 1 the end finally

Anyway, my way is more fun !

To find the real zeros of the polynomial function, we can use the rational zeros theorem. According to the rational zeros theorem, if a rational number p/q (where p is a factor of the constant term 28 and q is a factor of the leading coefficient 1) is a zero of the polynomial, then p must be a factor of 28 and q must be a factor of 1.

The factors of 28 are ±1, ±2, ±4, ±7, ±14, ±28, and the factors of 1 are ±1. So, the possible rational zeros are:
±1/1, ±2/1, ±4/1, ±7/1, ±14/1, ±28/1.

We can check these values by substituting them into the polynomial function and see if any of them yield a zero. Let's start with ±1/1:

For x = 1/1:
f(1/1) = (1/1)^4 + 6(1/1)^3 - 11(1/1)^2 - 24(1/1) + 28 = 1 + 6 - 11 - 24 + 28 = 0.

So, x = 1 is a zero of the polynomial.

Now, let's use polynomial division to reduce the polynomial by x - 1:

x^3 + 7x^2 - 18x + 28
--------------------------
x - 1 | x^4 + 6x^3 - 11x^2 - 24x + 28
-(x^4 - x^3)
----------------
7x^3 - 11x^2 - 24x + 28
-(7x^3 - 7x^2)
----------------
-4x^2 - 24x + 28
( -4x^2 + 4x)
----------------
-28x + 28
( -28x + 28)
----------------
0

After performing the polynomial division, we have:
f(x) = (x - 1)(x^3 + 7x^2 - 18x + 28)

Now, we can find the zeros of the cubic polynomial x^3 + 7x^2 - 18x + 28. To do this, we can use synthetic division or other methods. However, it turns out that this cubic polynomial does not have any rational zeros. Therefore, we cannot factor it further over the real numbers.

Thus, the real zeros of the original polynomial function f(x) = x^4 + 6x^3 - 11x^2 - 24x + 28 are x = 1, and the factored form over the real numbers is (x - 1)(x^3 + 7x^2 - 18x + 28).