Question

check my answers:

1. Which is equal to z3 8?
(Points : 1)
(z - 2)(z^2 - 2z - 4)

(z + 2)(z^2 - 2z + 4) <-----------this one

(z - 2)(z^2+ 2z + 4)

(z + 2)(z^2+ 2z - 4)

Answer 1

(z^3 -2z^2 - 4z) + ( 2z^2 -4z + 4)

z^3 -2z^2 + 2z^2 - 4z - 4z + 8

z^3 -8 + 8

so im lost

Answer 2

this is what i got: (z -2)(z2+ 2z + 4)

(z * z^2 + z * 2z + z * 4) – (-2 *z^2 – (-2) 2z – (-2) 4)
Z^3 + 2z^2 + 4z – 2z^2 -4z – 8
Z^3 + 2z^2 – 2z^2 + 4z – 4z – 8
Z^3 - 8

Answer 3

i chose c

Which polynomial is prime?
(Points : 1)
x3 + 49

x3 - 1000

x3 + 1

x3 - 64

Answer 4

and the cirst i chose c

Answer 5

i chose d

Which is equal to x2 + 2x + xy + 2y?
(Points : 1)
(x + y)(x - 2)

(x - y)(x - 2)

(x - y)(x + 2)

(x + y)(x + 2)




---------------------------------------------------------
(x + y)(x + 2):

X^2 + 2x + yx + 2y

Answer 6

Is it z3 - 8 i think you forgot the sign

Answer 7

yes it is i dont know why the (-) wont show up

Answer 8

You really need to represent exponentiation unabiguously. I'm going to guess that you mean

z^3 - 8

(z - 2) • (z^2 -2•z - 4)

(z + 2) • (z^2 -2•z + 4)

(z - 2) • (z^2+ 2•z + 4)

(z + 2) • (z^2+ 2•z - 4)

where "^" means exponentiation and "•" multiplication. In which case, the answer would be the third choice

(z - 2) • (z^2+ 2•z + 4)

which you wrote as

(z - 2)(z2+ 2z + 4)

Answer 9

yes

Answer 10

am i correct on my second answer?

Answer 11

the first one is correct, you need to stop canceling out your varible when subtracting. the only way you can take the variable out is if the coeficents add to zero

Answer 12

ok, when it comes to primes i was stuck between 1 and 49 (a, and c)

Answer 13

ok X³+1 can be factored to (x+1)(x²-x+1) but x³+49 can only be factored using complex numbers

Answer 14

ah i see, complex meaning decimals right?

Answer 15

no meaning square roots and imaginary numbers

Answer 16

oh ok, so i am correct on problem 2? x^3 - 1

Answer 17

no x³+1

Answer 18

i am sorry about that

Answer 19

wait no again that will factor the answer is the x³+49

Answer 20

i know that on problem 3 it was logical for my to choose d because:
3. Which is equal to x2 + 2x + xy + 2y?
(Points : 1)
(x + y)(x 2)

(x y)(x 2)

(x y)(x + 2)

(x + y)(x + 2)

they used (+) trough out the equation

Answer 21

that is a good way to narrow it down but then foil it to check that everything works

Answer 22

wait why? prime was divisiable by 1 and itself

Answer 23

because you can pull out an (x+1) and then manipulate the rest to make the equation equal the original by using (x²-x+1) multiply those 2 together and since they will = x³+1 that means it's not prime

Answer 24

you can't do that trick with the 49

Answer 25

i see...

Answer 26

what about problem 3?

Answer 27

was i correct on that one, and i need help with 2 others

Answer 28

yes

Answer 29

im stuck between a and d
4. Which is equal to x^3 - 2x^2 - 49x + 98?
(Points : 1)
(x - 7)(x + 7)(x + 2)

(x + 7)(x - 2)

(x2 + 49)(x - 2)

(x - 7)(x + 7)(x - 2)

Answer 30

ok for a and d you want to first multiple 2 together then multiply your answer to the third binomial

Answer 31

ok
:)

Answer 32

(x - 7)(x + 7)(x- 2)

X^2 + 7x – 7x -49
X^2 + x – 49
X^2 -49


(x^2 – 49 ) (x – 2)
X^3 – 2x^2 – 49x + 98

Answer 33

is this correct?

1. What is the solution set to the equation (x 4)(x + 3) = 0?
(Points : 1)
{3, -4}

{-3, -4}

{-3, 4}<-------------------this one

{3, 4}

Answer 34

yes

Answer 35

ok :)

Answer 36

What is the solution set to the equation a2 - 25 = 0?
(Points : 1)
{-5, 5} this one<-----------

{-5, -5}<----------------------this one too

{0, -5}

{0, 5}

Answer 37

i am going with a on that one

Answer 38

am i correct?