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

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0f(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
 one year ago
Best ResponseYou've already chosen the best response.0so at f(1) our function should look like this \[\LARGE f(1) =4(1)^8+3(1)^6+2(1)^4+2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@UsukiDoll so the answers 11?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0YEAH 1 to any exponent number will just be a number so \[\LARGE f(1) =4+3+2+2 =11\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could you help me with a few more please ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ack 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
 one year ago
Best ResponseYou've already chosen the best response.0thats fine , im a very patient person ! Given the function f(x)=x3−8x2+21x−18 what are the factors of of f(x) ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE f(x) =x^38x^2+21x18\] like this?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0hmmm...I wish I could use factor by grouping...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wish i was good at algebra :(

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ok 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.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0it's like setting that problem to 0 and breaking this function into at least 3 parts

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE 0 =x^38x^2+21x18\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0sigh... 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
 one year ago
Best ResponseYou've already chosen the best response.0so we need to plug those values in until we get our 0... so we need to let x =2, x = 3

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE f(2) =(2)^38(2)^2+21(2)18\] can you evaluate this function?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0the 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 qxp

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think so , is in 2?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yes 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 x2. That would give us the second polynomial we need

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437005051966:dw sorry it's hard to draw on a touchpad mouse

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so now we need to use long division . so let's start with dw:1437005170681:dw that portion only.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0what we need to do is to get rid of that x^3 so I have to multiply a variable with x2 so which variable do we need?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0what produces x^3 \[\large x(x^{?}) \rightarrow x^{1+?}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0similar to solveing 1 + x =3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a. (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
 one year ago
Best ResponseYou've already chosen the best response.0yes so we need to distribute (multiply) x^2 x^2(x2) which is x^32x^2 however we need to switch the signs when dividing with polynomials so x^3+2x^2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437005603335:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437005651594:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ok... so now I need to get rid of the 6x^2 and I have (x2) so which choice do we need. If I can get a copycat version of 6x^2 that would be great.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437005826577:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so right now I have x 2 what do I need to distribute to have 6x^2? I need to have at least one x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0[?] (x2) 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(x2)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0no I would have 3x^2 and our goal is to get rid of the 6x^2 similar to 7+7 =0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0well there is a 6 in the division. ?x(x2) perhaps using that number 6 would work 6x(x2) try distribute and tell me if you can get a 6x^2 ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0we also need a x^2 so \[x \cdot x =x^{1+1}\] what's 1+1?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yes so we have our x^2 x(x2) 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(x2) so using the distributive property 6x^2+12x switching the signs due to dividing polynomials 6x^212x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437006719993:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0at this point since all variables are used up, we just need a number

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0?(x2) what number do I need to produce a 9x?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0no.... 9 times ? = 9x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0OH MY I just typed it XD

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0we need a 9 9(x2) so that's 9x18 switching the signs 9x+18

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437007066892:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so our remainder is 10 but that's not important. What's important is that we have found the second polynomial which is x^26x+9 so from the rational root test we had x2 (x2)(x^26x+9) the second polynomial is obtained through long division can you factor (x^26x+9) ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0nice! (x2)(x3)^2 or (x2)(x3)(x3) now we can solve for x (x2)(x3)(x3) =0 since one factor is repeating we just have two cases x3=0 and x2=0 what is our x values for those 2 equations?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yes 3 and 2 so that is our roots.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the answers ( x  3)(x  2)(x  3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can 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
 one year ago
Best ResponseYou've already chosen the best response.0yes it is (x3)(x3)(x2).... ummm I really don't know how that next questino would work. .__.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay what about 2.Given the polynomial f(x)=6x5−4x3+4x+10 What one of the following is a possible rational root?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0we have to do the rational root test again so we find all factors of 10

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0try 1 and see if it works \[\LARGE f(1)=6(1)^54(1)^3+4(1)+10\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0if our result isn't 0, we need another root...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yeah time to toss that one out

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so 1 and 1 are fails. try 2 and 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it wasnt zero there either

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0then 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
 one year ago
Best ResponseYou've already chosen the best response.0that didnt work either !

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0right... none of the roots worked which means there aren't any with the rational root test.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay 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
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.0okay then , one more?Given the function f(x)=x3+10x2+31x+30 what one the following is a factor of f(x) ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE f(2) = (2)^3+10(2)^2+31(2)+30\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0woo HOO! which means we an divide that polynomial by qxp from the Factor theorem which is p/q=2/1 x(2) = x+2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437009518274:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so 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^32x^2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437009682298:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so now we need an 8x^2 so we need an 8x 8x(x+2) =8x^2+16x switching signs 8x^216x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437009861228:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.0can you factor x^2+8x+15?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0mhm 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
 one year ago
Best ResponseYou've already chosen the best response.0the answer was just (x+2) ! THANK YOU , you're a life saver!!
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