Question

Find ALL the zeros of g(x)=x^5+6x^4+13x^3+24x^2+36x . Part of the work must include synthetic division.

Answer 1

Answer:

$x=0,\:x=-3,\:x=-3,\:x=2i,\:x=-2i$

Step-by-step explanation:

$g(x)=x^5+6x^4+13x^3+24x^2+36x = x (x^4+6x^3+13x^2+24x+36)$

We have a zero at x = 0.

Possible rational zeros: (highest and lowest power's coefficients)

36 ⇒ 1, 2, 3, 4, 6, 9, 12, 18, 36

1 ⇒ 1

+ {1, 2, 3, 4, 6, 9, 12, 18, 36} and - {1, 2, 3, 4, 6, 9, 12, 18, 36}

f(-3) = 81 - 162 + 117 - 72 + 36 = 0

So, (x+3) is a factor.

Here comes the synthetic division part that I have shown in the attachment.

As shown in the attachment the function has double zeros at x = -3.

After synthetic division, the remaining part is $(x^2+4)$

It also means that we have 2 complex zeros at x = 2i and x=-2i

So the final form of g(x) is $g(x) = x(x+3)^2(x^2+4)$

Final solution is $x=0,\:x=-3,\:x=-3,\:x=2i,\:x=-2i$