write as product of linear factors? f(x) = x^4 + 2x^3 - 11x^2 + 8x - 60?

Question

just tell me how u got the answer like with synthetic dicvision or something

Answer 1

Any rational solutions will be factors of 60 (positive or negative). So we try a few to see if they make f(x) = 0. 3 works, so divide by x-3 (long or synthetic division doesn't matter, you get the same answer:)
f(x) = (x-3) (x^3 + 5x^2 + 4x + 20)
Similarly, x = -5 is a zero of this polynomial, so we get
f(x) = (x-3) (x+5) (x^2 + 4).
You could also get this by grouping the terms of the cubic appropriately.

From then it's simple: x^2 + 4 = (x - 2i) (x + 2i), so we have
f(x) = (x-3) (x+5) (x+2i) (x-2i).

Answer 2

just pick a number and see if the equation equals zero when it is subbed into the equation

lets say three

3^4 + 2*3^3 - 11*3^2 + 8*3 - 60
solving this i get 0

so one of your factors will be (x-3) --- just flip the sign of the number that you subbed in

now you need to do polynomial division


x^3+ 5x^2 + 4x + 20
____________________
(x-3) | x^4 + 2x^3 - 11x^2 + 8x - 60
--------x^4 - 3x^3
----------------5x^3 - 11x^2
---------------- 5x^3 - 15x^2
-------------------------- 4x^2 + 8x
-------------------------- 4x^2 - 12x
----------------------------------- 20x - 60
----------------------------------- 20x - 60
-------------------------------------------- 0

now repeat this process on the polynomial inside the bracket

until you get (x-3)(x+5)(x^2 +4)

but one of the factors are not linear so we need to use the

x^2 + 4 =0
x= (-4)^0.5
x=+2i and -2i
therefore
x-2i = 0 and x + 2i = 0
and x^2 + 4 = (x - 2i) (x + 2i)
so (x-3) (x+5) (x+2i) (x-2i)