Mathematics · Quantitative Aptitude

Algebraic Expressions and Polynomials

292 Questions

Algebraic expressions and polynomials form the foundation of algebra, involving variables, constants, and mathematical operations. This topic is heavily tested in the quantitative aptitude sections of SSC, banking, and state exams. Use these questions to practice expanding, factoring, and simplifying various polynomial expressions.

Expanding algebraic expressionsFactoring polynomialsAlgebraic identitiesFinding common factorsSimplifying numerical statements

Algebraic Expressions and Polynomials Questions

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths
If the expression $z^5 =32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5 - 32)=(z - 2) (z^2-pz+4)(z^2-qz+4)$, where p > q, then the value of $p^2-  2q$
  1. $8$
  2. $4$
  3. $-4$
  4. $-8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$ { z }^{ 5 }=32 $ can be factorized as
$(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)$ where $ p>q.$
 To find the value of $ { p }^{ 2 }-2q$
 Solution,
${ z }^{ 5 }=32\ { z }^{ 5 }-32=0$
$ { z }^{ 5 }-{ 2 }^{ 5 }=0$
$ z=2$ 
Will be one of the factor of given equation.
$ \quad \quad \quad \quad \quad \quad \quad { z }^{ 4 }+{ 2z }^{ 3 }+{ 4z }^{ 2 }+8z+16\ \therefore \quad z-2\sqrt { { z }^{ 5 }-32\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  } \ \quad \quad \quad \cfrac { \pm { z }^{ 5 }\mp { 2z }^{ 4 } }{ { \quad \quad \quad \quad \quad \quad 2z }^{ 4 }-32 } \ \quad \quad \quad \quad \quad \quad \cfrac { \pm { 2z }^{ 4 }\mp 4{ z }^{ 3 } }{ \quad \quad \quad \quad \quad \quad \quad 4{ z }^{ 3 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 4{ z }^{ 3 }\mp 8{ z }^{ 2 } }{ \quad \quad \quad \quad \quad \quad \quad 8{ z }^{ 2 }-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 8{ z }^{ 2 }\mp 16z }{ \quad \quad \quad \quad \quad \quad \quad 16z-32 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cfrac { \pm 16z\mp 32 }{ 0 } \ (z-2)({ z }^{ 4 }+2{ z }^{ 3 }+4{ z }^{ 2 }+8z+16)\longrightarrow (1)\ { z }^{ 5 }-32=(z-2)({ z }^{ 2 }-pz+4)({ z }^{ 2 }-qz+4)\quad \ { z }^{ 5 }-32=(z-2)({ z }^{ 4 }+(p+q)z^{ 3 }+(8+pq){ z }^{ 2 }+16-(4p+4q)z)\longrightarrow (2)$
On comparing $ (1)& (2)$ we get
$ p=-2-q\ 8+(-2-q)q=4\ -2q-{ q }^{ 2 }=-4\ { q }^{ 2 }+2q-4=0\ d=4+16=20\ q=\cfrac { -2\pm 2\sqrt { 5 }  }{ 5 } \ q=-1\pm \sqrt { 5 } \ { \parallel  }^{ rly }p=-1\pm \sqrt { 5 } $
But$\quad p>q\therefore p=-1+\sqrt { 5 } & q=-1-\sqrt { 5 } $
value of ${ p }^{ 2 }-2q={ \left( -1+\sqrt { 5 }  \right)  }^{ 2 }+2{ \left( -1-\sqrt { 5 }  \right)  }\ =1+5-2\sqrt { 5 } +2+2\sqrt { 5 } \ { p }^{ 2 }-2q=8\ $

Multiple choice maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

Fill in the blanks:
$10^2 +1^2 + 10^2 = 10^2$
$12^2 + 2^2 + 6^2 = 12^2$
$14^2 + 7^2$ + ____ = ____

  1. $3^2, 14^2$
  2. $2^2, 14^2$
  3. $2^2, 7^2$
  4. $2^2, 12^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From the pattern, the third number is the division of the first two number.
The fourth number can be obtained by same as the first number.
Then, the missing numbers will be
$14^2 + 7^2 + 2^2 = 14^2$
So, $2^2, 14^2$ are the missing numbers.

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Evaluate:
$2+2^2+2^3+....+2^9=$

  1. $1396$
  2. $1022$
  3. $1587$
  4. $1478$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$2+{ 2 }^{ 2 }+{ 2 }^{ 3 }+......+{ 2 }^{ 9 }$

The series is in $GP$ with common difference$=\cfrac { { 2 }^{ 2 } }{ 2 } =\cfrac { { 2 }^{ 3 } }{ 2 } .....=\cfrac { { 2 }^{ 9 } }{ { 2 }^{ 8 } } =2$
Sum of $GP=\cfrac { a({ r }^{ n }-1) }{ r-1 } $ where $a$ is the first term and $r$ is the common difference and last term$=a({ r }^{ n }-1)$
So,Last term,$2.{ 2 }^{ n-1 }={ 2 }^{ 9 }\ { 2 }^{ n-1 }={ 2 }^{ 8 }\ n-1=8\ n=9$
Sum$==\cfrac { 2({ 2 }^{ 9 }-1) }{ 2-1 } =2(512-1)\ =2(511)=1022$
Answer $(B)$.

Multiple choice business economics and quantitative methods measures of dispersion and skewness shortcut method to find variance and standard deviation variance and standard deviation measures of dispersion

Find $Var(2X+3)$

  1. $5 Var(X)+3$
  2. $4 Var(X)+3$
  3. $4 Var(X)$

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

$Var(2X+3)=Var(2X)+Var(3)$


$\implies Var(2X+3)=Var(2X)+0$ (Since, $Var(c)=0$)

$\implies Var(2X+3)=2^2Var(X)$ (Since, $Var(aX)=a^2Var(X)$)

$\implies Var(2X+3)=4Var(X)$

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

Find G.C.D of $20x^2-9x+1$ and $5x^2-6x+1$

  1. (x-1)

  2. (5x-1)

  3. (5x+1)

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

Let, $p(x) = 20x^2-9x+1$ and $q(x) = 5x^2-6x+1$
$p(x) = 20x^2-9x+1$
         $=20x^2-5x-4x+1$
         $=5x(4x-1)-1(4x-1)$
         $=(5x-1)(4x-1)$
and
$q(x) = 5x^2-6x+1$
         $=5x^2-5x-x+1$
         $=5x(x-1)-1(x-1)$
         $=(5x-1)(x-1)$
$\therefore$ G.C.D of $p(x)$  and  $q(x)=(5x-1)$.
Option B is correct.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

Find G.C.D of: $(x^2-9)(x-3)$ and $x^2+6x+9$

  1. $(x+3)^2$

  2. (x-3)

  3. (x+3)

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

$p(x) = (x^2-9)(x-3)$
        $= (x^2-3^2)(x-3)$
        $= (x-3)(x+3)(x-3)$
and
$q(x) = x^2+6x+9$
        $ = x^2+3x+3x+9$
        $ = x(x+3)+3(x+3)$
        $= (x+3)(x+3) $
$\therefore $ G.C.D of $p(x)$ and $q(x) = x+3 $
Option C is correct.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

$\begin{array}{l}\\left( {3x - 2y} \right)\left( {2x + y} \right)=\end{array}$

  1. $ =6{{x}^{2}}-xy+2{{y}^{2}}\,\, $
  2. $ =6{{x}^{2}}+xy-2{{y}^{2}}\,\, $
  3. $ =6{{x}^{2}}-xy-2{{y}^{2}}\,\, $
  4. $ =6{{x}^{2}}+xy+2{{y}^{2}}\,\, $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Part (1)

$ \left( a-2 \right)\left( a+2 \right) $

${{a}^{2}}-4$ 

Part (2)

$ \left( 3x-2y \right)\left( 2x+y \right) $

$ =6{{x}^{2}}+3xy-4xy-2{{y}^{2}} $

$ =6{{x}^{2}}-xy-2{{y}^{2}}\,\, $

Hence, this is the answer.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

lf the expression $3x^{2}+2pxy+2y^{2}+2ax-4y+1$ can be resolved into two linear factors, then $p$ must be a root of the equation

  1. $x^{2}+ax+6=0$
  2. $x^{2}+4ax+6=0$
  3. $x^{2}+4ax+2a^{2}+6=0$
  4. $x^{2}-4ax+6=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $3x^{2}+2pxy+2y^{2}+2ax-4y+1=0.$, then.
$\Delta =abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
$=3(2)(1)+2(-2)(a)(p)-3(-2)^{2}-2(a)^{2}-1(p)^{2}=0$
$6-4ap-12-2a^{2}-p^{2}=0$
$p^{2}+4ap+2a^{2}+6=0$
$\Rightarrow p $ is a solution of $x^{2}+4ax+2a^{2}+6=0$