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

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

Note by Swapnil Das
6 years, 1 month ago

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## Comments

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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

- 6 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 !

- 6 years, 1 month ago

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Extra marks for using calculus.

- 6 years, 1 month ago

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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!

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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Yeah!

- 6 years, 1 month ago

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Isileyeh tho thumeh itne doubts ah raheh hai!

- 6 years, 1 month ago

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Haan!

- 6 years, 1 month ago

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Nahi Milega.😐😑😑😑😑😑

- 6 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!

- 6 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).

- 6 years, 1 month ago

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Hi Swaplin, what are the answers?

- 6 years, 1 month ago

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

- 6 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 $\alpha$.

So $f(x)=(x-1)(x-1)(x-\alpha)=x^3-x^2(\alpha+2)+x(2\alpha+1)-\alpha$

As coefficient of $x^2$ must be zero so $\alpha=-2$. 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 $f(x)=(x-p)(x-q)=x^2-(p+q)x+pq$

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

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

Solve these simultaneous equations.

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

- 6 years, 1 month ago

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p=1,q=-2

- 6 years, 1 month ago

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Also p=q=0.

- 6 years, 1 month ago

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Yep! Missed that.

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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No need to @Mention!

- 6 years, 1 month ago

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You are welcome! :)

- 6 years, 1 month ago

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Don't fear when Brilliant.org is here!

- 6 years, 1 month ago

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Haha thanks!

- 6 years, 1 month ago

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Please reshare the note and help me! @satvik pandey

- 6 years, 1 month ago

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I guess that it is Swapnil LOL!

- 6 years, 1 month ago

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LOL! Swaplin :P :P

- 6 years, 1 month ago

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Lol

- 6 years, 1 month ago

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For the first question, we find that $\dfrac{f(x)}{(x-1)^{2}}=0$. 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, $x^3-3x+2$.

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

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

- 6 years, 1 month ago

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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 $x^{3}+bx+x=x^{3}-2x^{2}+x-zx^{2}+2xz-z$

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 $7+b=7-3=4$.

- 6 years, 1 month ago

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