Finding the zeros to the function f(x)=x^4 -x^3 -x^2 -x -2 HELP PLEASE?

Use synthetic division.

Best root to start guessing with is -1 (x + 1):

....____________

-1 | 1 -1 -1 -1 -2

........-1 +2..-1+2

......1.-2..1..-2..0

so -1 is a root and (x + 1) is a factor

now we have (x + 1)(x³ - 2x² + x - 2)

see if +2 is a root → (x - 2) is a factor:

....__________

+2 | 1 -2 +1 -2

........+2...0.+2

......1..0...1...0

so +2 is a root and (x - 2) is a factor

now we have (x + 1)(x - 2)(x² + 1) (the third binomial has imaginary roots)

x⁴ - x³ - x² - x - 2 = (x + 1)(x - 2)(x² + 1)

The function f(x) has two real roots; -1 and +2

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If you want to know how to get the imaginary roots, start with the quadratic formula:

[-b ± √(b² - 4ac)]/2a

[-(0) ± √(0)² - 4(1)(1)]/ 2(1) = ± [√-4]/2 = ±2[√-1]/2 = ± i

so the FULL factorization would be

(x + 1)(x - 2)(x + i)(x - i)

and all the zeros (real and imaginary) would be

-1, +2, +i, -i

set x^3 -x² - 2x to 0 you are able to ingredient an x out x(x² - x - 2) ingredient out what's in parenthesis x(x - 2)(x + a million) set each and each x to 0 x=0 x - 2 = 0 so x = 2 x + a million = 0 so x = -a million

Factored:

ƒ(x)=(x²+1)(x+1)(x-2)

x+1=0; then, x = -1

x-2=0; then, x = 2

x²+1=0; then, x² = -1; then, √x² = ±√-1;

(note that √-1 is equal to i); so,.

√x² = ±i; then, x = i, -i

x = 1,2,i,-i