Answered

What is a quartic function with only the two real zeros given?


x = -4 and x = -1

Ay = x^4 + 5x^3 + 5x^2 + 5x + 4
By = x^4 - 5x^3 - 5x^2 - 5x - 4
Cy = -x^4 + 5x^3 + 5x^2 + 5x + 4
Dy = x^4 + 5x^3 + 5x^2 + 5x - 5

Verified Answer :

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.

\(y = x^4 + 5x^3 + 5x^2 + 5x + 4\)

Put x=-1

\(y = (-1)^4 + 5(-1)^3 + 5(-1)^2 + 5(-1) + 4=0\)

Put x=-4

\(y = (-4)^4 + 5(-4)^3 + 5(-4)^2 + 5(-4) + 4=0\)

Therefore option A is correct.

\(y = x^4 - 5x^3 - 5x^2 - 5x - 4\)

The value of y is 2 at x=-1 and 512 at x=-4, therefore option B is incorrect.

\(y = -x^4 + 5x^3 + 5x^2 + 5x + 4\)

The value of y is -2 at x=-1 and -512 at x=-4, therefore option C is incorrect.

\(y = x^4 + 5x^3 + 5x^2 + 5x - 5\)

The value of y is -9 at x=-1 and -9 at x=-4, therefore option D is incorrect.

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