Question

factor of f(x)= x^4+8x^3+5x^2-38x+24 is

Answer 1

Do you know the "Rational Roots Theorem"?

Answer 2

no

Answer 3

If you have to factor something this complicated, you can be sure there are some simple numbers that will do the job!
Remember you are going to write it as: f(x) = a*b*c*d.
This means you can divide this polynomial by a, b, c, or d.
Also, setting f(x) = 0, this means a = 0 or etc.
Now when I try to factorize such a function, I put simple numbers like 1 or 2 in to see if this works.
INow setting x = 1 gives 0, which means:

f(x) = (x - 1) * (some 3rd degree polynomial)

We already have one factor!
The 3rd degree polynomial can be found by long division:\[x-1 \lceil x^4+8x^3+5x^2-38x+24 \rceil x^3\]Wait, this is impossible with the equation editor; see this pic:

Answer 4

It will help you to solve this, it is a lot to type. Very incompletely, you take the first and last coefficients and factor them taking each factor of the last coefficient and driving it by the first's. in this case x^4's coefficient factor(s) is 1, 24's are 1 ,2, 3, 4, 6, 8, 12, 24.
Divide each of 24's factors by x^4s and start subbing the smallest in for x to determine, if any cause the equation to equal zero.

x = (+/-)1/1 = +/-1,
x = (+/-)2/1 = +/-2. ...

If f(x) = 0 for a value then (x +/- value) can be divided into the initial equation.

A start?

Answer 5

"and dividing it by the first's" I meant to type.

Answer 6

I hope you understand the long division... in x.jpg.
Now\[f(x)=(x-1)(x^3+9x^2+14x-24)\]Believe it or not, you can divide by 1 another time!
Setting 1 in the right factor gives 1 +9 +14 - 24 = 0.
Doing another long division will crack it further:

Answer 7

Now we have\[f(x) = (x-1)^2(x^2+10x+24)\]The last part is of the 2nd degree.
Look for two numbers with sum 10 and product 24. These are 4 and 6.
So:\[f(x)=(x-1)^2(x+4)(x+6)\]

Answer 8

In my 2nd answer above, of course I divide by x - 1 another time, not by 1...