Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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

@UsukiDoll so the answers 11?

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

could you help me with a few more please ?

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

yes

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

i wish i was good at algebra :(

okay?

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

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

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

\[\LARGE f(2) =(2)^3-8(2)^2+21(2)-18\]
can you evaluate this function?

i think so , is in -2?

shouldn't be -2.

0?

|dw:1437005051966:dw| sorry it's hard to draw on a touchpad mouse

so now we need to use long division . so let's start with |dw:1437005170681:dw|
that portion only.

what produces x^3
\[\large x(x^{?}) \rightarrow x^{1+?}\]

similar to solveing 1 + x =3

|dw:1437005603335:dw|

|dw:1437005651594:dw|

|dw:1437005826577:dw|

so right now I have x -2
what do I need to distribute to have -6x^2?
I need to have at least one x

i dont know ?

-3?

no I would have -3x^2 and our goal is to get rid of the -6x^2
similar to -7+7 =0

i dont know :(

we also need a x^2 so \[x \cdot x =x^{1+1}\] what's 1+1?

|dw:1437006719993:dw|

at this point since all variables are used up, we just need a number

?(x-2) what number do I need to produce a 9x?

11? or 3?

no....
9 times ? = 9x

OH MY I just typed it XD

we need a 9
9(x-2) so that's 9x-18
switching the signs
-9x+18

|dw:1437007066892:dw|

(x-3)^2

3,2?

yes 3 and 2 so that is our roots.

so the answers ( x - 3)(x - 2)(x - 3)

yes it is (x-3)(x-3)(x-2).... ummm I really don't know how that next questino would work. .__.

we have to do the rational root test again so we find all factors of 10

try -1 and see if it works
\[\LARGE f(-1)=6(-1)^5-4(-1)^3+4(-1)+10\]

if our result isn't 0, we need another root...

i got 4

yeah time to toss that one out

so 1 and -1 are fails.
try 2 and -2

it wasnt zero there either

5 and -5

nope

that didnt work either !

right... none of the roots worked which means there aren't any with the rational root test.

try x=-2

\[\LARGE f(-2) = (-2)^3+10(-2)^2+31(-2)+30\]

i got 0!

|dw:1437009518274:dw|

|dw:1437009682298:dw|

so now we need an 8x^2
so we need an 8x
8x(x+2) =8x^2+16x
switching signs
-8x^2-16x

|dw:1437009861228:dw|

can you factor x^2+8x+15?

(x+3) (x+5)

mhm so now we have (x+2)(x+3)(x+5)
we can solve for x to obtain the roots
x+2=0
x+3=0
x+5=0

the answer was just (x+2) ! THANK YOU , you're a life saver!!

you're welcome :)