Polynomials & Real Numbers: Find Remainder & No. of Divisors

[sambarbarian]

this is a HOT question from polynomials .

1)find the remainder when x^100 is divided by x^2 - 3x + 2

another from real numbers

2) The number of + divisors which divide 10^999 but not 10^998 are ______ i tried them for sometime . but could not solve them .

 

[Curious3141]

this is a HOT question from polynomials .

1)find the remainder when x^100 is divided by x^2 - 3x + 2

another from real numbers

2) The number of + divisors which divide 10^999 but not 10^998 are ______


i tried them for sometime . but could not solve them .

What have you tried?

 

[sambarbarian]

well , i dint really know where to begin , so i just tried some factorization of the first equation and then dividing x^100 by them , but that dint work out . I also tried out a load of thinking but dint get anywhere :( . ( as to the second , I am clueless )

 

[Curious3141]

well , i dint really know where to begin , so i just tried some factorization of the first equation and then dividing x^100 by them , but that dint work out . I also tried out a load of thinking but dint get anywhere :( . ( as to the second , I am clueless )

Let's tackle the first, well, first.

One way to do this sort of problem is to use polynomial long division. While that may be practical for a lower order dividend (numerator), the dividend here is $x^{100}$, which makes this infeasible. So we have to use another approach.

The divisor here is a quadratic expression. Do you know what form the remainder will take? Would it be a constant (a number) like when you divide a polynomial by a linear divisor? Or something else?

Start by factorising the divisor ($x^2 - 3x + 2$).

 

[sambarbarian]

the form of remainder is not mentioned , and i tried the factorization , doesn't work

 

[Curious3141]

the form of remainder is not mentioned , and i tried the factorization , doesn't work

The remainder will be a polynomial expression. The order of the polynomial is at most one less than the order of the divisor. So if your divisor is a linear expression like x + 5, then you will have a constant remainder (like 3 or -5 or 0). If your divisor is a quadratic expression, your remainder may be a linear expression like x + 2 or 3x - 5, or it may be a constant like 2 or -5. All these are just examples to illustrate the point.

So you should start by assuming the remainder is a linear expression, ax + b, where a and b are constants. If a comes out to be 0, then you know the remainder is simply a constant.

When P(x) is divided by a quadratic expression $(x-\alpha)(x-\beta)$, you can express the result by this equation:

$P(x) = (x-\alpha)(x-\beta)Q(x) + (ax + b)$, where the remainder is $ax + b$.

So your first step is to find out $\alpha$ and $\beta$, which is why I asked you to factorise the quadratic divisor. Surely, you know how to do that!

 

[sambarbarian]

oh yes , i know that , i told you i tried that , but the equation cannot be solved :/

 

[Curious3141]

oh yes , i know that , i told you i tried that , but the equation cannot be solved :/

So WHAT is the factorisation of the divisor?

 

[Curious3141]

You say you know to factorise the quadratic divisor, and this is all I want to see. Once you do that, I can walk you through the rest of the solution. But you have to factorise the quadratic, which is a very basic step, which you say you know how to do.

So: just factorise $x^2 - 3x + 2$.

It's late where I am, so I'm going to have to turn in now. I'll check in again in the morning my time (about 8 hours' time) and help you, provided no one else has guided you to a solution by then.

 

[eumyang]

2) The number of + divisors which divide 10^999 but not 10^998 are ______

OP, start by writing the two numbers as products of a power of 2 and a power of 5. In other words,
$$10^{999} = 2^{(something)} \cdot 5^{(something)}$$

 

[LCKurtz]

Start by factorising the divisor ($x^2 - 3x + 2$).

... and i tried the factorization , doesn't work

If you can't factor that, why would you be expect to be able to solve this problem? Perhaps you should take a look in a high school algebra book and do a review.

 

[Curious3141]

If you can't factor that, why would you be expect to be able to solve this problem? Perhaps you should take a look in a high school algebra book and do a review.

Yes. Either he knows how to factor a quadratic, in which case he should show his work, or he doesn't, in which case this problem is not suitable for him at this stage.

 

[sambarbarian]

guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/

 

[sambarbarian]

OP, start by writing the two numbers as products of a power of 2 and a power of 5. In other words,
$$10^{999} = 2^{(something)} \cdot 5^{(something)}$$

u mean 2^999 * 3^999 ... ( why did you write 'something' ? )

 

[eumyang]

u mean 2^999 * 3^999 ... ( why did you write 'something' ? )

Because I wanted you to fill in (something) with a number. And by the way, the 2nd base should be 5, not 3. So,
$10^{999} = 2^{999} \cdot 5^{999}$ and
$10^{998} = 2^{998} \cdot 5^{998}$

You'll notice that 2⁹⁹⁹ is not a factor of 10⁹⁹⁸. In fact, any factor of 10⁹⁹⁹ that is divisible by 2⁹⁹⁹ cannot be a factor of 10⁹⁹⁸. So how many of these factors of 10⁹⁹⁹ would that make altogether? (And no, the answer to this question is not the final answer of the problem.)

 

[ehild]

guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/

First find the remainder when you divide x¹⁰⁰ by (x-1)._{(hint: is xⁿ-1 divisible by (x-1)?)}

ehild

 

[gabbagabbahey]

guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/

Take another look at Curious3141's post (#6). What do you get when you look at $x=\alpha$ and $x=\beta$? Apply that to your polynomial.

 

[Curious3141]

guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/

OK, finally, you've factored it (and shown it). (BTW, that's not an "equation", merely an expression).

You must have tried the wrong approach, because it's very easy from here on.

Use this equation I previously quoted:

$P(x) = (x-\alpha)(x-\beta)Q(x) + (ax + b)$, where the remainder is $ax + b$.

for the current problem as:

$x^{100} = (x-1)(x-2)Q(x) + (ax + b)$

and successively put $x = 1$ and $x = 2$ to get two simple, linear simultaneous equations in $a$ and $b$. Solve those, and just write down the remainder as $ax + b$, using what you've found.

 

[sambarbarian]

case 1 . 1^100 = (1)a + b => 1 = a+b
case 2 . 2^100 = (2)a = b => 2^100 = 2a+b

if i solve these for a and b , which value of x will i take for the answer ( 1 or 2 ) , as the remainder is ax+b.

this is the part where i failed.

 

[sambarbarian]

because i wanted you to fill in (something) with a number. And by the way, the 2nd base should be 5, not 3. So,
$10^{999} = 2^{999} \cdot 5^{999}$ and
$10^{998} = 2^{998} \cdot 5^{998}$

you'll notice that 2⁹⁹⁹ is not a factor of 10⁹⁹⁸. In fact, any factor of 10⁹⁹⁹ that is divisible by 2⁹⁹⁹ cannot be a factor of 10⁹⁹⁸. So how many of these factors of 10⁹⁹⁹ would that make altogether? (and no, the answer to this question is not the final answer of the problem.)

2^999 , 5^999 ?

 

[Curious3141]

case 1 . 1^100 = (1)a + b => 1 = a+b
case 2 . 2^100 = (2)a = b => 2^100 = 2a+b

Correct so far.

if i solve these for a and b , which value of x will i take for the answer ( 1 or 2 ) , as the remainder is ax+b.

this is the part where i failed.

No, the remainder IS (ax + b). The remainder when dividing by a quadratic can be a linear expression like that. You're probably getting confused by division by linear divisors, when the remainder is just a number (a constant).

I already explained this in post #6.

 

[sambarbarian]

by substitution , b=1-a => 2^100 = 1+a => a= 2^100 - 1
a=1-b =>2^100 = 2-b => b = 2 - 2^100

 

[Curious3141]

by substitution , b=1-a => 2^100 = 1+a => a= 2^100 - 1
a=1-b =>2^100 = 2-b => b = 2 - 2^100

Yes, now you can just write down the answer.

 

[sambarbarian]

x(2^100 - 1 ) + 2 - 2^100

2^100x -x +2 - 2^100 ? is this the ans ?

 

[Curious3141]

2^999 , 5^999 ?

No, I believe Eumyang was thinking about numbers like $2^{999}.5$, $2^{999}.5^4$ and $2^{999}.5^{999}$ (just examples). Have you considered those?

There are many more such numbers to consider.

 

[Curious3141]

x(2^100 - 1 ) + 2 - 2^100

2^100x -x +2 - 2^100 ? is this the ans ?

I would just leave it as $(2^{100} - 1 )x + 2 - 2^{100}$.

 

[sambarbarian]

any chance this is equal to 2 - x ,,, that's the answer at the back of the textbook

 

[Curious3141]

any chance this is equal to 2 - x ,,, that's the answer at the back of the textbook

No. How can it be? If $2 - x$ is the answer in your book, it's wrong.

 

[LCKurtz]

begin {offtopic}

Sambarbarian, I am curious to know if you grew up in or near northeastern South Dakota. If your answer is yes, I will tell you why I thought that. Otherwise nevermind.

end{offtopic}

 

[sambarbarian]

nopes , I am an indian ... but please tell me y u thought that