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 (x1)2{ (x-1) }^{ 2 } is a factor of f(x)=x3+bx+cf\left( x \right) ={ x }^{ 3 }+bx+c,then find the remainder when f(x)f\left( x \right) is divided by x2x-2

  • If (xp)(x-p) and (xq)(x-q) are the factors of x2+px+q{ x }^{ 2 }+px+q, then find the value of pp and qq.

Note by Swapnil Das
4 years, 1 month 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

Kalash Verma - 4 years, 1 month ago

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Bhai mere Karl mera polynomial chapter ka test hai. Agar main wahan Calculus use karunga to anda milega !

Swapnil Das - 4 years, 1 month ago

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Extra marks for using calculus.

Kalash Verma - 4 years, 1 month ago

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@Kalash Verma 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!

Swapnil Das - 4 years, 1 month ago

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@Swapnil Das What type of teacher is he? Doesn't believe in complex numbers!

Sravanth Chebrolu - 4 years, 1 month ago

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@Sravanth Chebrolu Yeah!

Swapnil Das - 4 years, 1 month ago

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@Swapnil Das Isileyeh tho thumeh itne doubts ah raheh hai!

Sravanth Chebrolu - 4 years, 1 month ago

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@Sravanth Chebrolu Haan!

Swapnil Das - 4 years, 1 month ago

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Nahi Milega.😐😑😑😑😑😑

Kalash Verma - 4 years, 1 month ago

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Well, do try using calculus and see the response of ur teacher. He'll be proud of u!

Aditya Kumar - 4 years, 1 month ago

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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

Rajdeep Dhingra - 4 years, 1 month ago

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Hi Swaplin, what are the answers?

satvik pandey - 4 years, 1 month ago

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Well I even donno, I wish someone could help, :(

Swapnil Das - 4 years, 1 month ago

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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\alpha.

So f(x)=(x1)(x1)(xα)=x3x2(α+2)+x(2α+1)αf(x)=(x-1)(x-1)(x-\alpha)=x^3-x^2(\alpha+2)+x(2\alpha+1)-\alpha

As coefficient of x2x^2 must be zero so α=2\alpha=-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)=(xp)(xq)=x2(p+q)x+pqf(x)=(x-p)(x-q)=x^2-(p+q)x+pq

So x2(p+q)x+pq=x2+px+qx^2-(p+q)x+pq=x^2+px+q

So (p+q)=p-(p+q)=p and pq=q pq=q

Solve these simultaneous equations.

Please check the calculation. I am not very good at it. :P

satvik pandey - 4 years, 1 month ago

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@Satvik Pandey p=1,q=-2

Kalash Verma - 4 years, 1 month ago

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@Kalash Verma Also p=q=0.

satvik pandey - 4 years, 1 month ago

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@Satvik Pandey Yep! Missed that.

Kalash Verma - 4 years, 1 month ago

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@Satvik Pandey I am grateful to you and @Sravanth Chebrolu for your help.

Swapnil Das - 4 years, 1 month ago

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@Swapnil Das No need to @Mention!

Sravanth Chebrolu - 4 years, 1 month ago

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@Swapnil Das You are welcome! :)

satvik pandey - 4 years, 1 month ago

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Don't fear when Brilliant.org is here!

Sravanth Chebrolu - 4 years, 1 month ago

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@Sravanth Chebrolu Haha thanks!

Swapnil Das - 4 years, 1 month ago

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Please reshare the note and help me! @satvik pandey

Swapnil Das - 4 years, 1 month ago

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I guess that it is Swapnil LOL!

Sravanth Chebrolu - 4 years, 1 month ago

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LOL! Swaplin :P :P

Nihar Mahajan - 4 years, 1 month ago

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Lol

Swapnil Das - 4 years, 1 month ago

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For the first question, we find that f(x)(x1)2=0\dfrac{f(x)}{(x-1)^{2}}=0. So dividing the polynomial, we find that:

  • For the remainder to be 00, we must have, (b1)x=4x(b-1)x= -4x and c=2c=2. Hence we find that b=(3)b=(-3) and c=2c=2.

  • So the polynomial will become, x33x+2x^3-3x+2.

  • Now, using remainder theorem, we find that the remainder of the polynomial is f(2)=23(3×2)+2=86+2=4f(2)=2^3-(3×2)+2 \\ = 8-6+2 \\= \boxed {4}

I just wanted to show the other way of doing this. Hope you found it useful!¨\huge \ddot\smile

Sravanth Chebrolu - 4 years, 1 month ago

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Remainder when we divide f(x)f(x) by (x1)2(x-1)^{2} is zero. Therefore 1=b+c-1=b+c.

Now remainder when f(x)f(x) is divided by x2x-2 is 8+2b+c8+2b+c i.e 7+b7+b.

Now we know by factor theorem that f(x)=(x1)2q(x)f(x)=(x-1)^{2}*q(x).

q(x)q(x) must be linear so let q(x)=xzq(x)=x-z.

Therefore f(x)=(x1)2(xz)f(x)=(x-1)^{2}(x-z)

Now x3+bx+x=x32x2+xzx2+2xzzx^{3}+bx+x=x^{3}-2x^{2}+x-zx^{2}+2xz-z

Comparing quadratic terms coef on both sides we get (2+z)=0-(2+z)=0 i.e z=2z=-2.

Now we compare linear terms we get b=1+2z=1+2(2)=14=3b=1+2z=1+2*(-2)=1-4=-3

Therefore our remainder is 7+b=73=47+b=7-3=4.

shivamani patil - 4 years, 1 month ago

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