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

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

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

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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 ?
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.
thats fine , im a very patient person ! Given the function f(x)=x3−8x2+21x−18 what are the factors of of f(x) ?
\[\LARGE f(x) =x^3-8x^2+21x-18\] like this?
yes
hmmm...I wish I could use factor by grouping...
i wish i was good at algebra :(
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.
okay?
it's like setting that problem to 0 and breaking this function into at least 3 parts
\[\LARGE 0 =x^3-8x^2+21x-18\]
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
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?
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
i think so , is in -2?
shouldn't be -2.
0?
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
|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 we need to do is to get rid of that x^3 so I have to multiply a variable with x-2 so which variable do we need?
what produces x^3 \[\large x(x^{?}) \rightarrow x^{1+?}\]
similar to solveing 1 + x =3
2
a. (x - 3)(x + 2)(x + 3) b.. (x + 3)(x + 2)(x + 3) c. (x + 3)(x - 2)(x - 3) d. (x - 3)(x - 2)(x - 3) these are the answer choices !!
yes so we need to distribute (multiply) x^2 x^2(x-2) which is x^3-2x^2 however we need to switch the signs when dividing with polynomials so -x^3+2x^2
|dw:1437005603335:dw|
|dw:1437005651594:dw|
ok... so now I need to get rid of the -6x^2 and I have (x-2) so which choice do we need. If I can get a copycat version of -6x^2 that would be great.
|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 ?
[?] (x-2) well we need to produce a -6x^2 so we need at least one x but what number do we need (and it has to be negative because we already have a negative number inside the divison and we want that to go) ?x(x-2)
-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 :(
well there is a -6 in the division. ?x(x-2) perhaps using that number -6 would work -6x(x-2) try distribute and tell me if you can get a -6x^2 ?
we also need a x^2 so \[x \cdot x =x^{1+1}\] what's 1+1?
2
yes so we have our x^2 x(x-2) when we distribute... but we also need a number so we can create -6x^2 as well... if we have -6 already in the division, then we need that same number -6x(x-2) so using the distributive property -6x^2+12x switching the signs due to dividing polynomials 6x^2-12x
|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|
so our remainder is 10 but that's not important. What's important is that we have found the second polynomial which is x^2-6x+9 so from the rational root test we had x-2 (x-2)(x^2-6x+9) the second polynomial is obtained through long division can you factor (x^2-6x+9) ?
(x-3)^2
nice! (x-2)(x-3)^2 or (x-2)(x-3)(x-3) now we can solve for x (x-2)(x-3)(x-3) =0 since one factor is repeating we just have two cases x-3=0 and x-2=0 what is our x values for those 2 equations?
3,2?
yes 3 and 2 so that is our roots.
so the answers ( x - 3)(x - 2)(x - 3)
can you help me with 4 or 5 more that is much easier than that one ? 1.A polynomial is to be constructed that has 5 turning points. What is the minimum degree of the polynomial?
yes it is (x-3)(x-3)(x-2).... ummm I really don't know how that next questino would work. .__.
okay what about 2.Given the polynomial f(x)=6x5−4x3+4x+10 What one of the following is a possible rational root?
we have to do the rational root test again so we find all factors of 10
so all factors of 10 is 1 2 5 10 but we also have to consider the negative versions as well so -1, -2, -5, -10 b
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
then the last two choices are -10 , 10 either one of them works or a math error happened. I do know that 1,-1 don't work at all
that didnt work either !
right... none of the roots worked which means there aren't any with the rational root test.
okay lets just skip that one , i have two more . Given the following polynomial find the maximum possible number of turning points. f(x)=x8+x14+x6+x13
well the answer to that previous question should be none because the rational root test failed and we have proven that by finding all roots and testing them out. And, again I don't know what a turning point is
okay then , one more?Given the function f(x)=x3+10x2+31x+30 what one the following is a factor of f(x) ?
the rational root test is used again and this time it's going to work all factors of 30 1 2 3 5 6 10 15 30 -1 -2 -3 -5 -6 -10 -15 -30
try x=-2
\[\LARGE f(-2) = (-2)^3+10(-2)^2+31(-2)+30\]
i got 0!
woo HOO! which means we an divide that polynomial by qx-p from the Factor theorem which is p/q=-2/1 x-(-2) = x+2
|dw:1437009518274:dw|
so we need to produce an x^3 but we have already done that earlier x^2(x+2) x^3+2x^2 switching signs -x^3-2x^2
|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|
well this works out nicely .. if we use the number 15 everything cancels out and there is no remainder so we have found our second polynomial (x+2)(x^2+8x+15)
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 :)

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