Tag: division algorithm for polynomials

Quizzes (0) Questions (64)

Questions Related to division algorithm for polynomials

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem 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. $3, 8$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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
         $f(2)=16-16+12-10+8=10$

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

$32x^{10}-33x^{5}+1$ is divisible by 

  1. $x-1$
  2. $x-2$
  3. $x-3$
  4. $x-4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(x-1)$

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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

  1. $4$
  2. $17$
  3. $23$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 .
Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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

  1. $1$
  2. $2$
  3. $-2$
  4. $-3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $ $\boxed{Always\, divisible\, by\, 9}$

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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. $2(1+x^2+x^4+....+x^{10})$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 ) $
Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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. $x$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Divide the polynomial by x^2+1 by substituting x^2 = -1. The terms become powers of (-1), simplifying to -x.

Multiple choice maths polynomials linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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

  1. $500$
  2. $501$
  3. $502$
  4. $499$
Reveal answer Fill a bubble to check yourself
A Correct answer
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!.