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# 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
1 year, 4 months 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 · 1 year, 4 months ago

Bhai mere Karl mera polynomial chapter ka test hai. Agar main wahan Calculus use karunga to anda milega ! · 1 year, 4 months ago

Extra marks for using calculus. · 1 year, 4 months 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! · 1 year, 4 months ago

What type of teacher is he? Doesn't believe in complex numbers! · 1 year, 4 months ago

Yeah! · 1 year, 4 months ago

Isileyeh tho thumeh itne doubts ah raheh hai! · 1 year, 4 months ago

Haan! · 1 year, 4 months ago

Well, do try using calculus and see the response of ur teacher. He'll be proud of u! · 1 year, 4 months ago

Nahi Milega.😐😑😑😑😑😑 · 1 year, 4 months ago

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 · 1 year, 4 months ago

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 $$7+b=7-3=4$$. · 1 year, 4 months ago

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!$$\huge \ddot\smile$$ · 1 year, 4 months ago

Hi Swaplin, what are the answers? · 1 year, 4 months ago

Well I even donno, I wish someone could help, :( · 1 year, 4 months ago

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 · 1 year, 4 months ago

p=1,q=-2 · 1 year, 4 months ago

Also p=q=0. · 1 year, 4 months ago

Yep! Missed that. · 1 year, 4 months ago

I am grateful to you and @Sravanth Chebrolu for your help. · 1 year, 4 months ago

You are welcome! :) · 1 year, 4 months ago

No need to @Mention! · 1 year, 4 months ago

Don't fear when Brilliant.org is here! · 1 year, 4 months ago

Haha thanks! · 1 year, 4 months ago

LOL! Swaplin :P :P · 1 year, 4 months ago

Lol · 1 year, 4 months ago

I guess that it is Swapnil LOL! · 1 year, 4 months ago