Question

Determine whether the given binomial is a factor of the polynomial P(x). Answer yes or no.

(x - 3); P(x) = 3x^4 + 3x^3 - 2x^2 - 2x

Answer 1

if (x-c) is a factor of P(x), then we could have the setup:\[\frac{(x-c)(other)}{(x-c)}\]
and at x=c we would get: 0/0

or in other words, if (x-c) is a factors, then P(c) = 0

Answer 2

3x^3(x+1)-2x(x+1)=(x+1)(3x^3-2x)
(x+1)*x*(3x^2-2)
no

Answer 3

can you help with a couple more?

Answer 4

yep

Answer 5

(x + 1); P(x) = 5x2 + 11x + 6

Answer 6

sorry. (x + 1); P(x) = 5x^2 + 11x + 6

Answer 7

do you know what it means to be a factor?

Answer 8

not in this sense

Answer 9

ignore the "sense"; how do you define a factor in general?

for example, what are the factors of 12?

Answer 10

1 and 12, 2 and 6, 3 and 4

Answer 11

good, then a factor is something that can be used to multiply and get the setup ... does that make sense?

Answer 12

if (x+1) is a factor of the given setup of P(x); then P(x) = (x+1) times something .... right?

Answer 13

yeah, so would you substitute the (x+1) for x?

Answer 14

not quite:

let P(x) = (x+1), times something; we should know that 0, times anything is equal to 0

would you agree then, that for x=-1; (-1+1), times something; is equal to zero?

Answer 15

yes, because -1+1=0

Answer 16

then ...

if P(-1) = 0, then (x+1) has to be a factor.

if P(-1) does NOT =0, then (x+1) cannot be a factor.

Answer 17

does: 5(-1)^2 + 11(-1) + 6 = 0 ?

Answer 18

20=0 so no?

Answer 19

5-11+6=0

Answer 20

but 5(-1)=-5and then you square it
-5*-5=25

so then you have 25 + 11(-1)

11(-1)=-11

25 - 11 = 14 and then ad 6

14+6 = 20

Answer 21

your squaring at the wring time:

\[5(-1)^2=5*-1*-1=-5*-1=5 \]

Answer 22

oh, Well okay! so then its 0=0

Answer 23

yes :) and since 0=0 we know that x+1 has to be a factor

Answer 24

so x+1 is a factor?

Answer 25

yes, since (x+1) = 0 when x=-1

and P(-1) = 0, then it has to be a factor

Answer 26

dbl chking the first one:

(x-3) ; P(x) = 3x^4 + 3x^3 - 2x^2 - 2x

since x-3 = 0, when x=3

P(3) = 3(3)^4 + 3(3)^3 - 2(3)^2 - 2(3)

3(3)^4 + 3(27) - 2(9) - 6

3(3)^4 + 3(27) - 24 ... there is no way this can be zero; therefore (x-3) cannot be a factor

Answer 27

Alright, I understand this a lot better now

Answer 28

good luck

Answer 29

Thanks, I am going to open another question to check my answers to some other questions could you help with that @amistre64?