Help me with 6 algebra 2 roots and zeros problems?

First question:

x^5 + 4x^3 = 0

The factoring is correct for the first step: x^3 (x^2 +4) = 0

To solve x^2 + 4 = 0...

x^2 = -4

x= (+/-) 2i (The "i" is the imaginary number).

Second equation: use the quadratic formula. This yields: [1 (+/-) sqrt(2)] / 4

Third and fourth questions:

Yes, use polynomial division to flush out one term.

Third question: I got x-3 to squeeze out.

(x-3)(x^2 - 2x - 1) = 0

The quadratic does not factor.

Answer: x={3, 1(+/-)sqrt(2)}

Fourth question: I got x+2 to squeeze out.

(x+2)(x^2 - x +3) = 0

Again, quadratic does not factor

Answer: x = {-2, [1(+/-)isqrt(11)]/2} (The "i" is the imaginary number).

Number 5----> ON YOUR OWN!! Sorry, I just don't know a way to do that one. Been too long since I did that, and no book to reference.

x^5 + 4x^3 = 0

x^3(x^2 + 4) = 0

the zero property of multiplication says

x^3 = 0

or

x^2 + 4 = 0

x^2 = -4

x = (+ or -) 2i

+++++++++++++++++++++++++++++++++++++++

h(x) = x^3 - 5x^2 + 5x + 3

use the rational root test, the only possible roots are (+ or -) 3

another way we could do this problem, would be...

given roots a, b, c

(x - a)(x - b)(x - c)

expands to

x^3 - (a + b + c)x^2 + (ab + bc + ac)x - abc

and thus we can form the system,

a + b + c = 5

ab + bc + ac = 5

abc = -3

+++++++++++++++++++++++++++++++++++++++

h(x) = x^3 + x^2 + x - 6

another rational root test, possible roots are:

(+ or -) 6

(+ or -) 3

(+ or -) 2

(+ or -) 1

or solve the system as described above.

++++++++++++++++++++

the key concept you are failing to recognize is that complex roots come in conjugate pairs...

thus,

if

- 5 + i

is a solution then

- 5 - i

is a solution as well.

Thus, multiplying the following out would give u the polynomial you're looking for...

(x + 3)(x + 5)(x - 1)(x - (-5 + i))(x - (-5 - i))

x^3(x^2 + 4) = 0

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

sorry this solution is completely wrong

(x^2-4)=(x-2)(x+2)

but (x^2-4) is not=(x+2)(x+2)