Question

35 POINT QUESTION, WILL MARK BRAINLIEST. Which of the following options is a 3rd degree polynomial with exactly 1 real root?

Answer 1

Answer:

A.  F(x) = x^3+9x^2+27x+27

Step-by-step explanation:

I am not 100% sure but I think this is right.

I found the root for this equation: x=-3

Hope this helps!

Answer 2

Answer:

The correct option is D.

Step-by-step explanation:

In option A,

The given function is

$F(x)=x^3+9x^2+27x+27$

$F(x)=x^3+3(3)x^2+3(3^2)x+3^3$

$F(x)=(x+3)^3$              $[\because (a+b)^3=a^3+3a^2b+3ab^2+b^3]$

Equate the function equal to zero, to find the roots.

$(x+3)(x-3)(x-3)=0\Rigtharrow x=-3$

The real root of this function is -3 with multiplicity 3. It means this function has 3 real roots.

In option B,

The given function is

$F(x)=x^3+3x^2-9x-27$

$F(x)=x^2(x+3)-9(x-3)$

$F(x)=(x^2-9)(x+3)$

$F(x)=(x+3)(x-3)(x+3)$         $[\because a^2-b^2=(a+b)(a-b)]$

Equate the function equal to zero, to find the roots.

$(x+3)(x-3)(x+3)=0\Rigtharrow x=-3,3,-3$

Therefore, this function has 3 real roots.

In option C,

The given function is

$F(x)=x^3-9x^2+27x-27$

$F(x)=x^3-3(3)x^2+3(3^2)x-3^3$

$F(x)=(x-3)^3$              $[\because (a-b)^3=a^3-3a^2b+3ab^2-b^3]$

Equate the function equal to zero, to find the roots.

$(x-3)(x-3)(x-3)=0\Rigtharrow x=3$

The real root of this function is 3 with multiplicity 3. It means this function has 3 real roots.

In option D,

$F(x)=x^3+3x^2+9x+27$

$F(x)=x^2(x+3)+9(x+3)$

$F(x)=(x+3)(x^2+9)$

Equate the function equal to zero, to find the roots.

$(x+3)(x^2+9)=0$

$x+3=0\Rightarrow x=-3$

$x^2+9=0\Rightarrow x^2=-9\Rightarrow x=\pm 3i(Imaginary)$

The roots of this functions are -3, 3i and -3i. Since this function has exactly one real root, therefore option D is correct.