Oh, so many questions!
Here are some of the questions of the chapter Polynomials that I am getting confused with. Please make some efforts answering my questions. Thanks!
If is a factor of ,then find the remainder when is divided by
If and are the factors of , then find the value of and .
Note by
Swapnil Das
6 years ago
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In Question 1 you can use the fact that when a polynomial have identical roots there differentiation also have the same root.Which means that f'(x) will also have (x-1) as a root. @Swapnil Das
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Bhai mere Karl mera polynomial chapter ka test hai. Agar main wahan Calculus use karunga to anda milega !
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Extra marks for using calculus.
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Are bhai hamara coaching teacher pagal Thai.WO sochta Thai ki "a polynomial of degree n can have maximum n roots, " air WO complex no.so mein bhibelievenahin Marta!
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What type of teacher is he? Doesn't believe in complex numbers!
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Yeah!
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Isileyeh tho thumeh itne doubts ah raheh hai!
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Haan!
Nahi Milega.😐😑😑😑😑😑
Well, do try using calculus and see the response of ur teacher. He'll be proud of u!
Since no one has posted the easiest solution I will.
We know the sum of roots is -b/a which in this case = 0.
We also know 2 roots are 1 hence the other root must be -2.
So the polynomial will be p(x) = (x-1)(x-1)(x+2).
@satvik pandey @Kalash Verma @Swapnil Das @Sravanth Chebrolu
Hi Swaplin, what are the answers?
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Well I even donno, I wish someone could help, :(
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The polynomial is of degree 3 so it should have three roots two of them are 1,1. Let the othe root be α.
So f(x)=(x−1)(x−1)(x−α)=x3−x2(α+2)+x(2α+1)−α
As coefficient of x2 must be zero so α=−2. Now using this you can form polynomial and then find remainder by remainder theorem.
2nd problem.
As the polynomial is of degree 2 so it should have 2 roots.
So f(x)=(x−p)(x−q)=x2−(p+q)x+pq
So x2−(p+q)x+pq=x2+px+q
So −(p+q)=p and pq=q
Solve these simultaneous equations.
Please check the calculation. I am not very good at it. :P
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p=1,q=-2
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Also p=q=0.
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Yep! Missed that.
I am grateful to you and @Sravanth Chebrolu for your help.
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No need to @Mention!
You are welcome! :)
Don't fear when Brilliant.org is here!
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Haha thanks!
Please reshare the note and help me! @satvik pandey
I guess that it is Swapnil LOL!
LOL! Swaplin :P :P
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Lol
For the first question, we find that (x−1)2f(x)=0. So dividing the polynomial, we find that:
For the remainder to be 0, we must have, (b−1)x=−4x and c=2. Hence we find that b=(−3) and c=2.
So the polynomial will become, x3−3x+2.
Now, using remainder theorem, we find that the remainder of the polynomial is f(2)=23−(3×2)+2=8−6+2=4
I just wanted to show the other way of doing this. Hope you found it useful!⌣¨
Remainder when we divide f(x) by (x−1)2 is zero. Therefore −1=b+c.
Now remainder when f(x) is divided by x−2 is 8+2b+c i.e 7+b.
Now we know by factor theorem that f(x)=(x−1)2∗q(x).
q(x) must be linear so let q(x)=x−z.
Therefore f(x)=(x−1)2(x−z)
Now x3+bx+x=x3−2x2+x−zx2+2xz−z
Comparing quadratic terms coef on both sides we get −(2+z)=0 i.e z=−2.
Now we compare linear terms we get b=1+2z=1+2∗(−2)=1−4=−3
Therefore our remainder is 7+b=7−3=4.