Respuesta :
A x+4=0 x+1=0 so factores (x+4)(x+1)=0; x^2+5x+4 is a factor. Divide it into x^4+5x^3+5x^2+4 and you get x^2+1 as the other factor which would be your other 2 roots (imaginary)
Answer:
The correct option is A.
Step-by-step explanation:
It is given that a quartic function has only two real zeros.
The degree of quartic function is 4. The function has only two real roots, therefore the roots has their multiplicity.
Put -1 and -4 for x. If the value of f(x)=0 at x=-4 and x=-1 then the function have two zeros -4 and -1.
[tex]y = x^4 + 5x^3 + 5x^2 + 5x + 4[/tex]
Put x=-1
[tex]y = (-1)^4 + 5(-1)^3 + 5(-1)^2 + 5(-1) + 4=0[/tex]
Put x=-4
[tex]y = (-4)^4 + 5(-4)^3 + 5(-4)^2 + 5(-4) + 4=0[/tex]
Therefore option A is correct.
[tex]y = x^4 - 5x^3 - 5x^2 - 5x - 4[/tex]
The value of y is 2 at x=-1 and 512 at x=-4, therefore option B is incorrect.
[tex]y = -x^4 + 5x^3 + 5x^2 + 5x + 4[/tex]
The value of y is -2 at x=-1 and -512 at x=-4, therefore option C is incorrect.
[tex]y = x^4 + 5x^3 + 5x^2 + 5x - 5[/tex]
The value of y is -9 at x=-1 and -9 at x=-4, therefore option D is incorrect.
