Question

5. What is a cubic polynomial function in standard form with zeros 1, –2, and 2? (1 point)
= x3 + x2 – 3x + 4
= x3 + x2 – 4x – 2
= x3 + x2 + 4x + 4
= x3 – x2 – 4x + 4

Answer 1

Cubic stands for ^3

Answer 2

Now, just factor each of them to find the zeros.

Answer 3

is the ansewer c

Answer 4

Just wait, let me check.

Answer 5

you can always use http://www.wolframalpha.com/ to help you.

Answer 6

y = x^3+x^2+4x+4

y = (1)^3+(1)^2+4(1)+4 ... plug in x = 1

y = 1 + 1 + 4(1) + 4

y = 1 + 1 + 4 + 4

y = 10

Since this is NOT zero, this means x = 1 is NOT a root or zero

Answer 7

so it can't be C

Answer 8

Oh, that's a simpler way to do it when you have the zeros given. Didn't see that :o

Answer 9

i need help i dont get it

Answer 10

The zeros are where the equation crosses on the x-axis on the graph.

Answer 11

The normal way (or the way I learnt it) is you have to factor your equation first so then you can find the zeros but @jim_thompson5910 's idea is way simpler.

Answer 12

if x = 1 is a zero, then plugging it into some equation *should* give you y = 0

Answer 13

ex:

if y = x-1

then x = 1 is a zero because

y = x-1

y = 1-1 .. plug in x = 1

y = 0

So this shows that x = 1 is a root or zero of y = x-1

Answer 14

would it be b then

Answer 15

what do you get when you plug x = 1 into y = x^3+x^2-4x-2

Answer 16

help
pleezzz

Answer 17

y = x^3+x^2-4x-2

y = (1)^3+(1)^2-4(1)-2

y = ???

Answer 18

y= -8

Answer 19

that is NOT y = 0, so choice B is also out

Answer 20

btw it should be y = -4, but that's still not 0

Answer 21

so what is the answer

Answer 22

well you've eliminated B and C so far

Answer 23

try out A

Answer 24

ok

Answer 25

is it a

Answer 26

what do you get when you plug in x = 1

Answer 27

i got 3

Answer 28

which is not 0, so x = 1 is not a root for choice A

Answer 29

so its d

Answer 30

yep

Answer 31

thank you

Answer 32

np