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 \(q\).

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TopNewestIn 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 · 2 years, 1 month ago

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– Swapnil Das · 2 years, 1 month ago

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

– Kalash Verma · 2 years, 1 month ago

Extra marks for using calculus.Log in to reply

– Swapnil Das · 2 years, 1 month ago

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!Log in to reply

– Sravanth Chebrolu · 2 years, 1 month ago

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

– Swapnil Das · 2 years, 1 month ago

Yeah!Log in to reply

– Sravanth Chebrolu · 2 years, 1 month ago

Isileyeh tho thumeh itne doubts ah raheh hai!Log in to reply

– Swapnil Das · 2 years, 1 month ago

Haan!Log in to reply

– Aditya Kumar · 2 years, 1 month ago

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

– Kalash Verma · 2 years, 1 month ago

Nahi Milega.😐😑😑😑😑😑Log in to reply

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 · 2 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 \(7+b=7-3=4\). – Shivamani Patil · 2 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!\(\huge \ddot\smile\) – Sravanth Chebrolu · 2 years, 1 month ago

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Hi Swaplin, what are the answers? – Satvik Pandey · 2 years, 1 month ago

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– Swapnil Das · 2 years, 1 month ago

Well I even donno, I wish someone could help, :(Log in to reply

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 – Satvik Pandey · 2 years, 1 month ago

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– Kalash Verma · 2 years, 1 month ago

p=1,q=-2Log in to reply

– Satvik Pandey · 2 years, 1 month ago

Also p=q=0.Log in to reply

– Kalash Verma · 2 years, 1 month ago

Yep! Missed that.Log in to reply

@Sravanth Chebrolu for your help. – Swapnil Das · 2 years, 1 month ago

I am grateful to you andLog in to reply

– Satvik Pandey · 2 years, 1 month ago

You are welcome! :)Log in to reply

– Sravanth Chebrolu · 2 years, 1 month ago

No need to @Mention!Log in to reply

Brilliant.orgis here! – Sravanth Chebrolu · 2 years, 1 month agoLog in to reply

– Swapnil Das · 2 years, 1 month ago

Haha thanks!Log in to reply

– Nihar Mahajan · 2 years, 1 month ago

LOL! Swaplin :P :PLog in to reply

– Swapnil Das · 2 years, 1 month ago

LolLog in to reply

nilLOL! – Sravanth Chebrolu · 2 years, 1 month agoLog in to reply

@satvik pandey – Swapnil Das · 2 years, 1 month ago

Please reshare the note and help me!Log in to reply