find all zeros of the function x^4+4x^3+4x^2-4x-5?

Answer 1

x^4+4x^3+4x^2-4x-5
= (x^4 - x^2) + (4x^3 - 4x) + (5x^2 - 5)
= x^2(x^2 -1) + 4x(x^2 -1) + 5(x^2 -1)
= (x^2 -1)(x^2 + 4x + 5)
= (x+1)(x-1)(x^2 + 4x + 5)
So, -1,1 and the roots of x^2 + 4x + 5 which are:

-4/2 ± √(16 - 20) /2 = -2 ± √(-4) /2 = -2 ± 2i/2 = -2 ± i

So,
1, -1, -2+i, -2-i

Answer 2

Let f(x)=x^4+4x^3+4x^2-4x-5

put x=1
=>f(1)=1+4+4-4+5=0
=>(x-1) is a factor.

put x=-1
=>f(-1)=1-4+4+4-5=0
=>(x+1) is a factor
=>f(x)=(x+1)(x-1){x^2 + 4x + 5}

Solving the quadratic:
x={-4(+-)sqrt(16-20)}/2
=>roots are imaginery

=>Only 2 points of zero
x=-1,1

Answer 3

the constant term is -5 its factors are 1,-1,5,-5

if it has a real root then it has to be one of

f(a) = 0

now f(1) = 1+4+4-4-5 = 0

so (x-1) is a factor or 1 is a zero

devide(siplit each trem so that combination with previous is divisible

x^4- x^3 + 5x^3 - 5x^2+ 9x^2 - 9x + 5x -5
= x^3(x-1) + 5x^2(x-1) + 9(x-1) +5(x-1)
= (x-1)(x^3+5x^2+9x+5)

let q(x) = x^3 + 5x^2+9x+ 5

the above does not have a positive root as all coefficeints are positive.

q(-1) = -1+5-9+5 =0

so -1 is another zero
so (x+1) is a factor

x^3+5x^2+9x +5 = (x^3+x^2+4x^2+4x+5x+5)
= (x+1)(x^2+4x+5)

now (x^2+4x+5) = (x^2+4x+4)+1

so (x+2)^2 = -1
x = 2 + i or 2 -i
the 4 zeroes are 1,-1, 2+i, 2-i

Answer 4

put x=1 &it satisfies the eq
so divide it by x-1
u get remainderx^3+5x^2+9x+5
now -1 satisfies it
so divide byx+1
u get remainderx^2+4x+5
x^2+4x+5can be solved by
discriminant method