Question

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

Answer 1

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 :)

Answer 2

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)

Answer 3

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)

Answer 4

I GET x^2 + 6 x -7
facor that
(x+7)(x-1)
so -7 and 1 the end finally

Answer 5

Anyway, my way is more fun !