Question

# Find all zeros of p(x), real or imaginary. p(x) = x^{4} + 6x^{3} + 6x^{2} -18x -27 List all of the possible rational zeros according to the rational z

Polynomial factorization
Find all zeros of p(x), real or imaginary. $$p(x) = x^{4} + 6x^{3} + 6x^{2} -18x -27$$ List all of the possible rational zeros according to the rational zero theorem and state the values for C, A, B and D in the following partial factorization of $$p(x) = (x-c)(x^{3}+Ax^{2}+Bx+D)$$ State the exact answer and a decimal approximation of each zero to the tenths place

2021-05-02
Step 1
Zeros of a polynomial:
$$p(x)=x^{4}+6x^{3}+6x^{2}-18x-27$$
To find zeros of a polynomial we need to put p(x)=0 and then simplify the polynomial in factor form.
Step 2
$$x^{4}+6x^{3}+6x^{2}-18x-27=0$$
$$x^{4}+3x^{3}+3x^{3}+9x^{2}-3x^{2}-9x-9x-27=0$$
$$x^{3}(x+3)+3x^{2}(x+3)-3x(x+3)-9(x+3)=0$$
$$(x+3)(x^{3}+3x^{2}-3x-9)=0$$
$$(x+3)(x^{2}(x+3)-3(x+3))=0$$
$$(x+3)(x+3)(x^{2}-3)=0$$
$$(x+3)(x-3)(x+\sqrt{3})(x-\sqrt{3})=0$$
The zeros are (x+3)=0. x=-3
(x-3)=0. x=3
$$(x+\sqrt{3})=0. x=-\sqrt{3}$$
$$(x-\sqrt{3})=0. x=\sqrt{3}$$
The zeros are $$x = 3,-3,\sqrt{3},-\sqrt{3}$$
=3,-3,1.7,-1.7
The values of A,B,C,D are 3,-3,-3,-9 respectively.