Tag: multiplication and division of algebraic expressions

Questions Related to multiplication and division of algebraic expressions

The remainder obtained when the polynomial $1+x+x^ {3}+x^ {9}+x^ {27}+x^ {81}+x^ {243}$ is divisible by $x-1$ is

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

  1. $3$

  2. $5$

  3. $7$

  4. $11$

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Correct Option: C
Explanation:

$p{\left( x \right)} = 1 + x + {x}^{3} + {x}^{9} + {x}^{27} + {x}^{81} + {x}^{243}$

Let $q{\left( x \right)}$ be the quotient when $P{\left( x \right)}$ divided by $\left( x - 1 \right)$.
Therefore,
$P{\left( x \right)} = \left( x - 1 \right) \cdot q{\left( x \right)} + A$
$P{\left( 1 \right)} = \left( 1 - 1 \right) \cdot q {\left( x \right)} + A$
$7 = 0 + A$
$A = 7$
Hence the remainder is $7$.

If $A=2x^{3}+5x^{2}+4x+1$ and $B=2x^{2}+3x+1$, then find the quotient from the following four option, when A is divided by B.

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

  1. $x-1$

  2. $x+1$

  3. $2x+1$

  4. $2x-1$

Show answer
Correct Option: B
Explanation:


$\dfrac{A}{B} = \dfrac{2x^{3}+5x^{2}+4x+1}{2x^{2}+3x+1}$

$=\dfrac{2x^{3}+(3x^{2}+2x^{2})+(x+3x)+1}{2x^{2}+2x+1}$

$=\dfrac{(2x^{3}+3x^{2}+x)+(2x^{2}+3x+1)}{2x^{2}+3x+1}$

$=\dfrac{(2x^{2}+3x+1)(x+1)}{2x^{2}+3x+1}=x+1$

Evaluate: $\displaystyle \frac{a^3\, +\, b^3\, +\, c^3\, -\, 3abc}{a^2\, +\, b^2\, +\, c^2\, -\, ab\, -\, bc\, -\, ca}$

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

  1. $0$

  2. $a + b + c$

  3. $1$

  4. None of these

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Correct Option: B
Explanation:

$\cfrac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca } \ =\cfrac { \left( a+b+c \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right)  }{ { (a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca) } \ =a+b+c$.

If x + 2 and x-1 are the factors of $x^3 + 10x^2+mx + n$, then the values of m and n are respectively

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

  1. 5 and -3

  2. 17 and -8

  3. 7 and-18

  4. 23 and -19

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Correct Option: C
Explanation:

Here, $x + 2$ is a factor of $x^3 + 10x^2 + mx + n$
$x =-2$
$(-2)^3 + 10(-2)^2 + m(-2) + n =0$
$ -8 + 40 = 2m - n $
$2m -n = 32$                     .....(i)
Again, $x-1 $ is a factor of $x^3 + 10x^2 + mx + n$
$x =1$
$1+10+m+n=0$
$m + n =-11$                    .....(ii)
Adding (i) and (ii). we get,
$3m = 21$
$m=7$
By putting m in (i). we get,
$2(7) - n = 32 $
$ - n = 18 $
   $n = -18$
Option C is correct.

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.

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

  1. $-3, 8$

  2. $3,-8$

  3. $-3,-8$

  4. $3, 8$

Show answer
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?

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

  1. $-2$

  2. $-1$

  3. $1$

  4. $2$

Show answer
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 $(x-2)$

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

  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

Show answer
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 

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

  1. $x-1$

  2. $x-2$

  3. $x-3$

  4. $x-4$

Show answer
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 

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

  1. $4$

  2. $17$

  3. $23$

  4. $0$

Show answer
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'

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

  1. $1$

  2. $2$

  3. $-2$

  4. $-3$

Show answer
Correct Option: A
Explanation:

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

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