Tag: division algorithm for polynomials

Questions Related to division algorithm for polynomials

The expression $2x^3 + ax^2 + bx +3$, where a and b are constants, has a factor of x-1 and leaves a remainder of 15 when divided by x+2. Find the value of a and b respectively.

  1. $-3, 8$

  2. $3,-8$

  3. $-3,-8$

  4. $3, 8$


Correct Option: B
Explanation:

f(x)=$ 2x^3+ax^2-bx+3$ 
At x=2 
f(2)=15 
f(1)=0 
f(x)=$ 2x^3+ax^2-bx+3$ 
f(1)= 2+a-b+3=0 
a-b+5=0......A 
f(x)= $2x^3+ax^2-bx+3$ 
f(2)=$ 2(2^3)+a(2^2)-2b+3=15 $
4a-2b=-4 
Multiply A by 2 and subtract from above equation 
4a-2b=-4 
2a-2b+10=0 
2a-10=-4 
2a= 6 
a=3 
From A 
3-b+5=0 
8-b=0 
b=8 
So a=3 and b=8

If $(x^{3} + 5x^{2} + 10k)$ leaves remainder $-2x$ when divided by $(x^{2} + 2)$, then what is the value of k?

  1. $-2$

  2. $-1$

  3. $1$

  4. $2$


Correct Option: C
Explanation:

$x^{3} + 5x^{2} + 10k$
$= (x^{2} + 2)(x + 5) + 10k - 2x - 10$
$\Rightarrow 10k - 2x - 10 = -2x$
$\Rightarrow 10k - 10 = 0$ or $k = 1$.

The polynomial $f(x)={ x }^{ 4 }-2{ x }^{ 3 }+3{ x }^{ 2 }-ax+b$ when divided by $(x-1)$ and $(x+1)$ leaves the remainders $5$ and $19$ respectively. Find the values of $a$ and $b$. Hence, find the remainder when $f(x)$ is divided by $(x-2)$

  1. $a=6,b=8$ and $remainder=10$

  2. $a=5,b=8$ and $remainder=10$

  3. $a=5,b=7$ and $remainder=10$

  4. None of these


Correct Option: B
Explanation:

    by remainder theorem,

      $f(1)=5$ $and$ $f(-1)=14$

     $\therefore 1-2+3-a+b=5\Rightarrow b-a=3$  and
     $1+2+3+a+b=19\Rightarrow a+b=13$ 
      $b=8$ $a=5$ 
       and remainder when $f(x)$ is divided by $(x-2)$ is
         $f(2)=16-16+12-10+8=10$

$32x^{10}-33x^{5}+1$ is divisible by 

  1. $x-1$

  2. $x-2$

  3. $x-3$

  4. $x-4$


Correct Option: A
Explanation:
$32x^{10}- 33 x^5 +1=0$
Let $m = x^5$
$\Rightarrow 32 m^2 - 33 m + 1 =0$
$\Rightarrow (32 m -1)(m-1)=0$
$\Rightarrow (32x^5-1)(x^5-1)=0$
$\therefore 32x^{10}- 33 x^5 +1=(32x^5-1)(x^5-1)$

$x^n-y^n$ is always divisible by $(x-y)$

$\therefore \,32x^{10}- 33 x^5 +1$ is divisible by $(x-1)$

The remainder, when $({ 15 }^{ 23 }+{ 23 }^{ 23 })$ is divided by $19$, is 

  1. $4$

  2. $17$

  3. $23$

  4. $0$


Correct Option: D
Explanation:
Given, $(15)^{23}+(23)^{23}$

$=(19-4)^{23}+(19+4)^{23}$

$\Rightarrow $ In bino  expansion  of above expression the term containing  19 well be cancelled 
as they disable by 19 then remaining term are 

$\Rightarrow  (-4)^{23}+(4)^{23}=0$

Therefore the remainder is exactly  zero .

If $(x^{100} + 2x^{99} + K)$ is exactly divisible by $(x + 1)$, find the value of 'K'

  1. $1$

  2. $2$

  3. $-2$

  4. $-3$


Correct Option: A
Explanation:

$x^{100}+2x^{99}+k$ is exactly divisible by $(x+1)$

$\therefore x=-1$ is the root of $x^{100}+2x^{99}+k$
$\Rightarrow (-1)^{100}+2(-1)^{99}+k=0$ 
$\Rightarrow 1-2+k=0$ 
$\Rightarrow \boxed{k=1}$

The sum of the digits of a 3 digit number is subtracted from the number. The resulting number is always.

  1. Divisible by 6

  2. Not divisible by 6

  3. Divisible by 9

  4. Not divisible by 9


Correct Option: C
Explanation:

Let the no. be $xyz$ sum of digit is $(x+y+z)$ 

as $xyz=100x+10y+z$
then $xyz-(x+y+z)=99x+9y$  
$\therefore $ $\boxed{Always\, divisible\, by\, 9}$

The remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+....+x^{11}$ is?

  1. $0$

  2. $2$

  3. $1+x^2+x^4+...+x^{10}$

  4. $2(1+x^2+x^4+....+x^{10})$


Correct Option: D
Explanation:
$ \left( \sum _{n=0}^N x^{n} \right) = \left ( \dfrac{x^{N+1}-1}{x-1} \right ) $

$ \Rightarrow Dividend = \left ( \dfrac{x^{24}-1}{x^{2}-1} \right ) $
$ Divisor = \left ( \dfrac{x^{12}-1}{x-1} \right ) $

Now,
$ \left ( \dfrac{x^{24}-1}{x^{2}-1} \right )  = \left ( \dfrac{x^{12}-1}{x-1} \right ) \left ( \dfrac{x^{12}-1+2}{x+1} \right ) = \left ( \dfrac{\left ( x^{12}-1 \right )^{2}}{x^{2}-1} \right ) + 2\left ( \dfrac{x^{12}-1}{x^{2}-1} \right ) $

$ \Rightarrow Remainder = 2\left ( 1+x^{2}+x^{4}...+x^{10} \right ) $

If the polynomial $x^{19}+x^{17}+x^{13}+x^{11}+x^7+x^5+x^3$ is divided by $(x^2+1)$, then the remainder is:

  1. $1$

  2. $x^2+4$

  3. $-x$

  4. $x$


Correct Option: C

The decimal representation of $2005!$ ends with $m$ zeroes then $m=$

  1. $500$

  2. $501$

  3. $502$

  4. $499$


Correct Option: A
Explanation:
We know that a number gets a zero at the end of it if the number has 10 as a factor.
  
So I need to find out how many times 10 is a factor in the expansion of 23!.

But since 5×2 = 10, I need to account for all the products of 5 and 2.
 Looking at the factors in the above expansion, there are many more numbers that are multiples of 2 (2, 4, 6, 8, 10, 12, 14,...) than are multiples of 5 (5, 10, 15,...). 
That is, if I take all the numbers with 5 as a factor, I'll have way more than enough even numbers to pair with them to get factors of 10 (and another trailing zero on my factorial).
No. of multiples of 5 between 1 and 2005!= 401
No. of multiples of 25= 80
no. of multiples of 125=16
no. of multiples of 625=3
Hence total multiples of 5 and hence 10 are 500.
There are 500 zeroes in the end of 2005!.