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

 

[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}

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

Because I did grow up in South Dakota, and I have met people from the eastern part of the state that use the word "dint" when they mean "didn't", which is a contraction for "did not". The word "dint" means something else.

 

[eumyang]

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.)

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

There are many more such numbers to consider.

Right. 2⁹⁹⁹ (or 2⁹⁹⁹ x 5⁰) is not a factor of 10⁹⁹⁸.
2⁹⁹⁹ x 5 (or 2⁹⁹⁹ x 5¹) is not a factor of 10⁹⁹⁸.
2⁹⁹⁹ x 5² is not a factor of 10⁹⁹⁸.
These are some of the factors of 10⁹⁹⁹, divisible by 2⁹⁹⁹, but are not factors of 10⁹⁹⁸. So how many factors would this make altogether?

 

[Curious3141]

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

Well, hello, I'm (ethnically) Indian too, although I'm a Singaporean by birth and citizenship.

And as LCKurtz points out, "dint" has its own meaning. It means by virtue of (something) when used in a phrase like "by dint of his dogged hard work". It's usually used emphatically, to suggest force, power or perseverance. It can also be an alternative spelling for "dent".

Back to the math now!

 

[sambarbarian]

right. 2⁹⁹⁹ (or 2⁹⁹⁹ x 5⁰) is not a factor of 10⁹⁹⁸.
2⁹⁹⁹ x 5 (or 2⁹⁹⁹ x 5¹) is not a factor of 10⁹⁹⁸.
2⁹⁹⁹ x 5² is not a factor of 10⁹⁹⁸.
These are some of the factors of 10⁹⁹⁹, divisible by 2⁹⁹⁹, but are not factors of 10⁹⁹⁸. So how many factors would this make altogether?

999 + 999 = 1998 ?

 

[Curious3141]

999 + 999 = 1998 ?

I have one more than that. How did you arrive at your figure (exactly)?