Quantitative Aptitude · Mathematics

Number and Polynomial Division

133 Questions

Number and polynomial division tests your skill in finding quotients, remainders, and applying theorems to algebraic expressions. These calculations are a staple in quantitative aptitude exams. Solving these problems enhances speed and accuracy for competitive tests.

Number remaindersPolynomial divisionQuotient and remainderFactor and remainder theoremSuccessive division

Number and Polynomial Division Questions

Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

The quotient when  $1 + x ^ { 2 } + x ^ { 4 } + x ^ { 6 } + \dots + x ^ { 34 }$  is divided by  $1 + x + x ^ { 2 } + x ^ { 3 } + \dots + x ^ { 17 }$  is

  1. $x ^ { 17 } - x ^ { 15 } + x ^ { 13 } - x ^ { 11 } + \dots + x$
  2. $x ^ { 17 } + x ^ { 15 } + x ^ { 13 } + x ^ { 11 } + \dots + x$
  3. $x ^ { 17 } + x ^ { 16 } + x ^ { 15 } + x ^ { 14 } + \dots + 1$
  4. $x ^ { 17 } - x ^ { 16 } + x ^ { 15 } - x ^ { 14 } + \dots - 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$1+x^2+x^4+\cdots+x^{34}=\dfrac{1-x^{36}}{1-x^2}$

$1+x+x^2+x^3+\cdots+x^{17}=\dfrac{1-x^{18}}{1-x}$

$\implies \dfrac{1+x^2+x^4+x^6+\cdots +x^{34}}{1+x+x^2+\cdots+x^{17}}=\dfrac{1-x^{36}}{1-x^{18}}\times \dfrac{1-x}{1-x^2}=\dfrac{1+x^{18}}{1+x}=x^{17}-x^{16}+x^{15}+\cdots-1$
Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

On dividing 55,390 by 299 the remainder is 75. The quotient is

  1. 195

  2. 185

  3. 175

  4. 193
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
It is given that the dividend is $55390$, divisor is $299$ and the remainder is $75$.

Let the quotient be $x$.

We know that 

Dividend = Divisor × quotient + remainder

Therefore, by substituting the values, we have,

$55390=299\times x+75$
 $\Rightarrow 55390=299x+75$

$\Rightarrow 299x=55390-75$

$\Rightarrow 299x=55315$

$\Rightarrow x=\dfrac { 55315 }{ 299 }$

$\Rightarrow x=185$

Hence, the quotient is $185$.
Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

The least number which on division by 35 leaves a remainder 25 and on division by 45 leaves the remainder 35 and on division by 55 leaves the remainder of 45 is

  1. 2515

  2. 3455

  3. 2875

  4. 2785
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$35=5\times7$
$45=3\times3\times5$
$55=5\times11$
LCM of (35,45,55)=$3\times3\times5\times\times7\times11=3465$
Since difference between divisor and remainder is 10
Hence least number is $3465-10=3455$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If ${x}^{3}+m{x}^{2}+nx+6$ has $(x-2)$ as factor and leaves a remainder $3$ when divided by $(x-3)$ find the values of $m,\,n$

  1. $m=2,n=2$
  2. $m=2,n=-2$
  3. $m=-2,n=1$
  4. $m=-3,n=-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$x-2$ is factor 


$\Rightarrow x=2$
$f\left( 2 \right) =14+4m+2n$
Remainder is zero

$\Rightarrow 7+2m+n=0\quad \longrightarrow \left( i \right) $
Now, $x-3=0$
gives remainder $3$

$\Rightarrow f\left( 3 \right) =3$
$\Rightarrow 33+9m+3n=3$
$\Rightarrow 10+3m+n=0\quad \longrightarrow \left( ii \right) $

From $(i)$ & $(ii)$
$m=-3$
$n=-1$

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

If quotient = $3x^2\, -\, 2x\, +\, 1$, remainder = $2x - 5$ and divisor  = $x + 2$, then the dividend is:

  1. $3x^3\, -\,4x^2\, +\, x\, -\, 3$
  2. $3x^3\, -\, 4x^2\, -\, x\, +\, 3$
  3. $3x^3\, +\, 4x^2\, -\, x\, +\, 3$
  4. $3x^3\, +\, 4x^2\, -\, x\, -\,3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Dividend = (divisor $\times$ quotient) + remainder
= $(3x^2\, -\, 2x\, +\, 1)\, \times\,  (x\, +\, 2)\,+\, (2x\, -\, 5)$
= $3x^3\, +\, 4x^2\, -\, x\,-\, 3$

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

The remainder if $a{x}^{3}+b{x}^{2}+cx+d$ is divided by $ax+b$

  1. $ad-bc$
  2. $\cfrac{1}{a}(ad-bc)$
  3. $\cfrac{a-bc}{d}$
  4. $\cfrac{a+b+cd}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $p(x) = a{x}^{3}+b{x}^{2}+cx+d$ and factor of $ax+b$ is
$ax+b = 0$
$x = -\frac ba$
$ p(-\frac ba) = a(-\frac ba)^{3}+b(-\frac ba)^{2}+c(-\frac ba)+d$
$p(-\frac ba) = -\frac {b^3}{a^2}+ \frac {b^3}{a^2} -\frac {bc}{a}+d $
$p(-\frac ba) =  -\frac {bc}{a}+d $
$p(-\frac ba) = \frac 1a (ad - bc) $
when $a{x}^{3}+b{x}^{2}+cx+d$ is divided by $ax+b$ then remainder is $p(-\frac ba) = \frac 1a (ad - bc) $
Option B is correct.

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

There is a remainder of 3 when a number is divided by 6. What will be the remainder if the square of the same number is divided by 6?

  1. 1

  2. 0

  3. 3

  4. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the no. be $'x'$ $\Rightarrow x=6q+3$

$x^{2}=(6q+3)^{2}=36q^{2}+36q+9$
Divide $x^{2}$ by $6$ 
we get $\boxed{x^{2}=6(6q^2+6q+1)+3}$
Hence the remainder is $\boxed{3}$

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

Find the reminder when ${x^3} + 3{x^2} + 3x + 1$ is divided by $x + \pi $

  1. $\pi$
  2. $-{\pi}^{2}+3{\pi}^{3}-3\pi+1$
  3. $-{\pi}^{3}+3{\pi}^{2}-3\pi+1$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The remainder theorem states that when a polynomial, $f\left(x\right)$, is divided by a linear polynomial , $x - a$, the remainder of that division will be equivalent to $f\left(a\right)$.

Given:$f\left(x\right)={x}^{3}+3{x}^{2}+3x+1$

$f\left(x\right)$ is divided by a linear polynomial , $x+\pi$, the remainder of that division will be equivalent to $f\left(-\pi\right)$.

Remainder$=f\left(-\pi\right)={\left(-\pi\right)}^{3}+3{\left(-\pi\right)}^{2}+3\left(-\pi\right)+1=-{\pi}^{3}+3{\pi}^{2}-3\pi+1$
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

Find the quotient $q(x)$ and remainder $r(x)$ of the following when $f(x)$ is divided by $g(x)$.
$p(x)=x^3-3x^2-x+3$;
$g(x)=x^2-4x+3$

  1. $q(x)=x-1$ and $r(x)=0$
  2. $q(x)=x+1$ and $r(x)=0$
  3. $q(x)=2x+1$ and $r(x)=0$
  4. $q(x)=2x-1$ and $r(x)=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider the polynomial $f(x)=x^3-3x^2-x+3$ and factorise it as follows:


$f(x)=x^3-3x^2-x+3=x^2(x-3)-1(x-3)=(x^2-1)(x+3)=(x+1)(x-1)(x-3)$

Therefore, $f(x)=(x+1)(x-1)(x-3)$

Now consider the polynomial $g(x)=x^2-4x+3$ and factorise it as follows:

$g(x)=x^2-4x+3=x^2-3x-x+3=x(x-3)-1(x-3)=(x-1)(x-3)$

Therefore, $g(x)=(x-1)(x-3)$


Now divide $f(x)$ by $g(x)$ to get $q(x)$:

$q(x)=\frac { (x-1)(x+1)(x-3) }{ (x-1)(x-3) } =x+1$

Since $f(x)$ is divisible by $g(x)$, therefore, the remainder $r(x)=0$.

Hence, the quotient $q(x)=x+1$ and remainder $r(x)=0$.

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

Find the quotient $q(x)$ and remainder $r(x)$ of the following when $f(x)$ is divided by $g(x)$.
$p(x)=x^6+x^4-x^2-1$;
$g(x)=x^3-x^2+x-1$

  1. $q(x)=2x^3+x^2+x-1$ and $r(x)=0$
  2. $q(x)=x^3+x^2+x-1$ and $r(x)=0$
  3. $q(x)=2x^3+x^3+x+1$ and $r(x)=0$
  4. $q(x)=x^3+x^2+x+1$ and $r(x)=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Consider the polynomial $f(x)=x^6+x^4-x^2-1$ and factorise it as follows:


$f(x)=x^6+x^4-x^2-1=x^4(x^2+1)-1(x^2+1)=(x^4-1)(x^2+1)=(x^2-1)(x^2+1)(x^2+1)$
$=(x-1)(x+1)(x^2+1)^2$

Therefore, $f(x)=(x-1)(x+1)(x^2+1)^2$

Now consider the polynomial $g(x)=x^3-x^2+x-1$ and factorise it as follows:

$g(x)=x^3-x^2+x-1=x^2(x-1)+1(x-1)=(x^2+1)(x-1)$

Therefore, $g(x)=(x-1)(x^2+1)$


Now divide $f(x)$ by $g(x)$ to get $q(x)$:

$q(x)=\frac { (x-1)(x+1)(x^{ 2 }+1)^{ 2 } }{ (x-1)(x^{ 2 }+1) } =(x+1)(x^{ 2 }+1)=x^{ 3 }+x^{ 2 }+x+1$

Since $f(x)$ is divisible by $g(x)$, therefore, the remainder $r(x)=0$.

Hence, the quotient $q(x)=x^{ 3 }+x^{ 2 }+x+1$ and remainder $r(x)=0$.


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

On dividing $f(x)=2x^5+3x^4+4x^3+4x^2+3x+2$ by a polynomial $g(x)$, where $g(x)=x^3+x^2+x+1$, the quotient obtained as $2x^2+x+1$. Find the remainder $r(x)$.

  1. $r(x)=7x^3+x^2-1$
  2. $r(x)=3x^2+2x+1$
  3. $r(x)=x-2$
  4. $r(x)=x+1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By remainder theorem,
$f(x)=q(x)g(x)+r(x)$
$\therefore 2x^5+3x^4+4x^3+4x^2+3x+2=(2x^2+x+1)(x^3+x^2+x+1)+r(x)$
$=2x^2(x^3+x^2+x+1)+x(x^3+x^2+x+1)+1(x^3+x^2+x+1)+r(x)$
$=2x^5+2x^4+2x^3+2x^2+x^4+x^3+x^2+x+x^3+x^2+x+1+r(x)$
$=2x^5+3x^4+4x^3+4x^2+2x+1+r(x)$
$r(x)=x+1$

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

For a polynomial, dividend is $\displaystyle x^{4}+4x-2x^{2}+x^{3}-10$, quotient is $\displaystyle x^{2}+3x-3x^{2}+4x+12$ and remainder is $14$, then divisor is equal to

  1. $\displaystyle x^{2}+2$
  2. $\displaystyle x^{2}-2$
  3. $\displaystyle x+2$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p(x)=x^4+4x-2x^2+x^3-10$ and $q(x)=x^2+3x-3x^2+4x+12$ and $r(x)=14$

$p(x)-14$ is not divisible by $(x^2+2),(x^2-2),(x+2)$  
hence correct option is (D)