Question

given the function f(x)=4x^8+3x^6+2x^4+2 what is the value of f(1)

Answer 1

f(x) =f(1) means that our x =1 so we plug in x =1 throughout the function
\[\LARGE f(x) =4x^8+3x^6+2x^4+2\]

Answer 2

so at f(1) our function should look like this
\[\LARGE f(1) =4(1)^8+3(1)^6+2(1)^4+2 \]

Answer 3

@UsukiDoll so the answers 11?

Answer 4

YEAH 1 to any exponent number will just be a number so
\[\LARGE f(1) =4+3+2+2 =11\]

Answer 5

could you help me with a few more please ?

Answer 6

ack one to any exponent number will just be 1 so then we have 4(1) which is 4 3(1) which is 3 2(1) which is 2
si 4+3+2+2 is 11.
I could, but I am on a netbook, and it's a bit buggy, so you'll have to be a bit patient with me.

Answer 7

thats fine , im a very patient person !
Given the function f(x)=x3−8x2+21x−18
what are the factors of of f(x) ?

Answer 8

\[\LARGE f(x) =x^3-8x^2+21x-18\] like this?

Answer 9

yes

Answer 10

hmmm...I wish I could use factor by grouping...

Answer 11

i wish i was good at algebra :(

Answer 12

ok factor by grouping is out of the question. I got to use something that I haven't done in years. Using the rational root theorem.

Answer 13

okay?

Answer 14

it's like setting that problem to 0 and breaking this function into at least 3 parts

Answer 15

\[\LARGE 0 =x^3-8x^2+21x-18\]

Answer 16

sigh... ok... we have to find all factors of 18 which
is 1 2 3 6 9 18
and each of them has + or - signs so we got a bunch of choices to choose from
1 2 3 6 9 18
-1 -2 -3 -6 -9 -18

Answer 17

so we need to plug those values in until we get our 0... so we need to let x =2, x = 3

Answer 18

\[\LARGE f(2) =(2)^3-8(2)^2+21(2)-18\]

can you evaluate this function?

Answer 19

the reason why I ask this is that we need an x that can produce a 0 and then afterwards we have to use the Factor theorem which states that if p/q is that root in the polynomial then it can be divided by qx-p

Answer 20

i think so , is in -2?

Answer 21

shouldn't be -2.

Answer 22

0?

Answer 23

yes so now that we got our x or in that theorem p/q
which is 2/1 (yeah I forgot to write the roots in fraction form), Then it's possible to divide that original function with x-2. That would give us the second polynomial we need

Answer 24

sorry it's hard to draw on a touchpad mouse