anonymous
  • anonymous
given the function f(x)=4x^8+3x^6+2x^4+2 what is the value of f(1)
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
f(x) =f(1) means that our x =1 so we plug in x =1 throughout the function \[\LARGE f(x) =4x^8+3x^6+2x^4+2\]
UsukiDoll
  • UsukiDoll
so at f(1) our function should look like this \[\LARGE f(1) =4(1)^8+3(1)^6+2(1)^4+2 \]
anonymous
  • anonymous
@UsukiDoll so the answers 11?

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UsukiDoll
  • UsukiDoll
YEAH 1 to any exponent number will just be a number so \[\LARGE f(1) =4+3+2+2 =11\]
anonymous
  • anonymous
could you help me with a few more please ?
UsukiDoll
  • UsukiDoll
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.
anonymous
  • anonymous
thats fine , im a very patient person ! Given the function f(x)=x3−8x2+21x−18 what are the factors of of f(x) ?
UsukiDoll
  • UsukiDoll
\[\LARGE f(x) =x^3-8x^2+21x-18\] like this?
anonymous
  • anonymous
yes
UsukiDoll
  • UsukiDoll
hmmm...I wish I could use factor by grouping...
anonymous
  • anonymous
i wish i was good at algebra :(
UsukiDoll
  • UsukiDoll
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.
anonymous
  • anonymous
okay?
UsukiDoll
  • UsukiDoll
it's like setting that problem to 0 and breaking this function into at least 3 parts
UsukiDoll
  • UsukiDoll
\[\LARGE 0 =x^3-8x^2+21x-18\]
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
so we need to plug those values in until we get our 0... so we need to let x =2, x = 3
UsukiDoll
  • UsukiDoll
\[\LARGE f(2) =(2)^3-8(2)^2+21(2)-18\] can you evaluate this function?
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
i think so , is in -2?
UsukiDoll
  • UsukiDoll
shouldn't be -2.
anonymous
  • anonymous
0?
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
|dw:1437005051966:dw| sorry it's hard to draw on a touchpad mouse
UsukiDoll
  • UsukiDoll
so now we need to use long division . so let's start with |dw:1437005170681:dw| that portion only.
UsukiDoll
  • UsukiDoll
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?
UsukiDoll
  • UsukiDoll
what produces x^3 \[\large x(x^{?}) \rightarrow x^{1+?}\]
UsukiDoll
  • UsukiDoll
similar to solveing 1 + x =3
anonymous
  • anonymous
2
anonymous
  • anonymous
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 !!
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
|dw:1437005603335:dw|
UsukiDoll
  • UsukiDoll
|dw:1437005651594:dw|
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
|dw:1437005826577:dw|
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
i dont know ?
UsukiDoll
  • UsukiDoll
[?] (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)
anonymous
  • anonymous
-3?
UsukiDoll
  • UsukiDoll
no I would have -3x^2 and our goal is to get rid of the -6x^2 similar to -7+7 =0
anonymous
  • anonymous
i dont know :(
UsukiDoll
  • UsukiDoll
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 ?
UsukiDoll
  • UsukiDoll
we also need a x^2 so \[x \cdot x =x^{1+1}\] what's 1+1?
anonymous
  • anonymous
2
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
|dw:1437006719993:dw|
UsukiDoll
  • UsukiDoll
at this point since all variables are used up, we just need a number
UsukiDoll
  • UsukiDoll
?(x-2) what number do I need to produce a 9x?
anonymous
  • anonymous
11? or 3?
UsukiDoll
  • UsukiDoll
no.... 9 times ? = 9x
UsukiDoll
  • UsukiDoll
OH MY I just typed it XD
UsukiDoll
  • UsukiDoll
we need a 9 9(x-2) so that's 9x-18 switching the signs -9x+18
UsukiDoll
  • UsukiDoll
|dw:1437007066892:dw|
UsukiDoll
  • UsukiDoll
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) ?
anonymous
  • anonymous
(x-3)^2
UsukiDoll
  • UsukiDoll
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?
anonymous
  • anonymous
3,2?
UsukiDoll
  • UsukiDoll
yes 3 and 2 so that is our roots.
anonymous
  • anonymous
so the answers ( x - 3)(x - 2)(x - 3)
anonymous
  • anonymous
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?
UsukiDoll
  • UsukiDoll
yes it is (x-3)(x-3)(x-2).... ummm I really don't know how that next questino would work. .__.
anonymous
  • anonymous
okay what about 2.Given the polynomial f(x)=6x5−4x3+4x+10 What one of the following is a possible rational root?
UsukiDoll
  • UsukiDoll
we have to do the rational root test again so we find all factors of 10
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
try -1 and see if it works \[\LARGE f(-1)=6(-1)^5-4(-1)^3+4(-1)+10\]
UsukiDoll
  • UsukiDoll
if our result isn't 0, we need another root...
anonymous
  • anonymous
i got 4
UsukiDoll
  • UsukiDoll
yeah time to toss that one out
UsukiDoll
  • UsukiDoll
so 1 and -1 are fails. try 2 and -2
anonymous
  • anonymous
it wasnt zero there either
UsukiDoll
  • UsukiDoll
5 and -5
anonymous
  • anonymous
nope
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
that didnt work either !
UsukiDoll
  • UsukiDoll
right... none of the roots worked which means there aren't any with the rational root test.
anonymous
  • anonymous
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
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
okay then , one more?Given the function f(x)=x3+10x2+31x+30 what one the following is a factor of f(x) ?
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
try x=-2
UsukiDoll
  • UsukiDoll
\[\LARGE f(-2) = (-2)^3+10(-2)^2+31(-2)+30\]
anonymous
  • anonymous
i got 0!
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
|dw:1437009518274:dw|
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
|dw:1437009682298:dw|
UsukiDoll
  • UsukiDoll
so now we need an 8x^2 so we need an 8x 8x(x+2) =8x^2+16x switching signs -8x^2-16x
UsukiDoll
  • UsukiDoll
|dw:1437009861228:dw|
UsukiDoll
  • UsukiDoll
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)
UsukiDoll
  • UsukiDoll
can you factor x^2+8x+15?
anonymous
  • anonymous
(x+3) (x+5)
UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
the answer was just (x+2) ! THANK YOU , you're a life saver!!
UsukiDoll
  • UsukiDoll
you're welcome :)

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