Finding zeros of a quartic function

[DTA]

Homework Statement

The problem states:

Consider the function f(x)=x⁴-8x³+2x²+80x-75

a) Verify that x-1 and x-5 are factors
b) Find the remaining factors of f(x)
c) List all real zeros

Homework Equations

I did synthetic division to prove that 1 and 5 are factors, yet I'm having trouble figuring out how to get the remaining zeros.

The Attempt at a Solution

I tried splitting up the equation and factoring out from both sides and got:

x³(x-8)+2x(x-40)-75

And I just got lost from here. I'm blanking on how to find the zeros besides just choosing a bunch of numbers and doing synthetic division a bunch of times.

Thanks for any help!

 

[Tedjn]

If x-1 and x-5 are factors, what do f(1) and f(5) equal, and why?

 

[vela]

You might also find it helpful to use the results of the synthetic division to get a polynomial that's easier to factor.

 

[DTA]

They both equal zero. I already understood that part, but my teacher is asking for any remaining factors of f(x). I used my graphing calculator and saw f(-3) is another zero, but I don't know how to factor f(x) to algebraically find f(-3) as a zero.

 

[Srumix]

Since you know that x-1 and x-5 are factors of the function f(x)=x^4-8x^3+2x^2+80x-75 then there exist a function, call it p(x) which as the property that if multiplied with the two factors it yields the function f(x). In other words:

(Assuming x-1 and x-5 are factors)

f(x) = x⁴ - 8x³ + 2x² + 80x - 75
->
f(x) = p(x) * (x-1) * (x-5)

You will be able to find p(x) by dividing f(x) with (x-1)(x-5).

Now you utilize that p(x)'s factors are also factors of the 'original' function, f(x). And you get the factors for p(x), obviously, from solving the equation p(x) = 0.

(I'm very tired and English is not my native language, so I hope I was able to provide you with some hints)

 

[vela]

Synthetic division gives you both a remainder and a quotient. Say you verified the x-5 case first. The bottom row of numbers you got would be 1, -3, -13, 15, 0. The final zero tells you that x-5 divided evenly into f(x); the first four numbers give you the quotient. So you have $f(x)=(x-5)(x^3-3x^2-13x+15)$. Now if you do synthetic division of $x^3-3x^2-13x+15$ by x-1, you'll be left with a quadratic, which you can hopefully factor by trial and error or by inspection.