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 maths square and square root scientific notation use of exponents power of 10

Evaluate the product of $2.3\times 10^4$ and $3\times 10^3$.

  1. $6.9\times 10^4$
  2. $6.9\times 10^7$
  3. $6.9\times 10^8$
  4. $6.9\times 10^6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(2.3\times 10^4) \times (3 \times 10^3)$
$\Rightarrow (2.3 \times 3) \times(10^{3+4})$

$\therefore 6.9 \times 10^7$
Ans- Option $B$.

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

Which of the following represents the given expression?
$a^2b^3\times 2ab^2$ ?

  1. $2a^3b^4$
  2. $2a^3b^5$
  3. $2ab$
  4. $a^3b^5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We know that if a number say $b$ is multiplied three times. That is, $b\times b\times b$ can be written as $b^3$.
=> $b^{3}$ = $b\times b\times b$
Similarly,
$a^{2}b^3$ = $a \times a\times b\times b\times b$
$2ab^{2}$ = $ 2 \times a\times b\times b$.
Thus, 
$a^{2}b^{3}\times 2ab^{2} = a \times a\times b\times b\times b \times 2 \times a\times b\times b $.
                      $= 2 \times a \times a \times a\times b\times b\times b\times b\times b$.
                      $ = 2a^{3}b^{5}$.
Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

Let $x _{1}, x _{2}, .....x _{n}$ be in an AP of $x _{1} + x _{4} + x _{9} + x _{11} + x _{20} + x _{22} + x _{27} + x _{30} = 272$, then $x _{1} + x _{2} + x _{3} + ..... + x _{30}$ is equal to

  1. $1020$
  2. $1200$
  3. $716$
  4. $2720$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If an AP consist of $30$ terms, Then $x _{1} + x _{30} = x _{4} + x _{27} = x _{9} + x _{22} = x _{11} + x _{20}$
$\because x _{1} +x _{4} + x _{9} + x _{11} + x _{20} + x _{27} + x _{30} = 272$
$\Rightarrow (x _{1} + x _{30}) + (x _{4} + x _{27}) + (x _{9} + x _{22}) + (x _{11} + x _{26}) = 272$
$\Rightarrow 4(x _{1} + x _{30}) = 272$
$\Rightarrow x _{1} + x _{30} = \dfrac {272}{4} = 68$
$S _{30} = \dfrac {30}{2} (x _{1} + x _{30}) = 15\times 68 = 1020$

Multiple choice reciprocal equations theory of equations maths

lf $\mathrm{f}({x})=0$ is a reciprocal equation of second type and even degree, then a factor of $\mathrm{f}({x})$  is:

  1. $x+1$
  2. $x-1$
  3. $x^{2}-1$
  4. $x^{2}$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

The equation will have 2 solutions, 1 and -1.
The solution will be x+1 and x-1,
$(x-1)(x+1)=x^2-1$

Multiple choice reciprocal equations theory of equations maths

lf $\mathrm{f}(\mathrm{x})=0$ is a reciprocal equation of first type and odd degree, then a factor of $\mathrm{f}(\mathrm{x})$ is:

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

When the reciprocal equation is of an odd degree, $x=-1$ is always a solution.
So, $x+1$ is a factor.

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

$(9p - 5q)^{2} + 180 pq$ is equivalent to _______.

  1. $(5p + 9q)^{2}$
  2. $(5p - 9q)^{2}$
  3. $(9p + 5q)^{2}$
  4. $(9p - 5q)^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know that $a^2+b^2+2ab$
$(a-b)^2=a^2+b^2-2ab$
$\Rightarrow (a-b)^2+4ab=a^2+b^2-2ab+4ab=(a+b)^2$    $(\because a=9p$ and $b=5q)$
$\Rightarrow 4ab=4(9p)(5q)=180pq$
Therefore, $ (9p-5q)^2+180pq=(9p+5q)^2$

Option (C) is correct.
Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

A factor of $(3x^{4} - 12y^{4})$ is _________.

  1. $3$
  2. $x^{2} - 2y^{2}$
  3. $x^{2} + 2y^{2}$

  4. All of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given expression is $(3x^{4}-12y^{4})$
Taking $3$ common, we get
$= 3(x^{4}-4y^{4})$
$= 3(x^4-(\sqrt{2}y)^{4})$
We know the identity and applying it $a^{2}-b^{2} = (a-b)(a+b)$

$3x^{4}-12y^{4}= 3(x^{4}-(\sqrt{2}y)^{4})$

$3x^{2}-12y^{2}= 3(x^{2}-\sqrt{2}y^{2})(x^{2}+\sqrt{2}y^{2})$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

Evaluating the following :
$(1+2\sqrt{x})^{5}+(1-2\sqrt{x})^{5}$

  1. $2(1+40x^2+80x)$
  2. $2(1-40x+81x^2)$
  3. $2(1+40x+80x^2)$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given to evaluate is $(1+2\sqrt x)^5 +(1-2\sqrt x)^5$

$\Rightarrow 2[^{5}C _{0} (2\sqrt x)^0 +^5C _2 (2\sqrt x)^2 +^5C _4 (2\sqrt x)^4]$

$\Rightarrow 2[1+10\times 4x+5\times 16x^2]$

$\Rightarrow 2[1+40x+80x^2]$

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Which of the following quadratics is irreducible?

  1. $2x^2 - 5x + 3$
  2. $2x^2 - 5x - 3$
  3. $5x^2 - 2x + 3$
  4. $5x^2 - 2x - 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can check by comparing the D of quadratic equations with their relation with 0


$2x^2-5x+3=0$
$D=(-5)^2-4(2)(3)=25-24=1>0$
Reducible

$2x^2-5x-3=0$
$D=(-5)^2-4(2)(-3)=25+24=49>0$
Reducible

$5x^2-2x+3=0$
$D=(-2)^2-4(5)(3)=4-60=-56<0$
Irreducible


$5x^2-2x-3=0$
$D=(-2)^2-4(5)(-3)=4+60=64>0$
Rreducible


Therefore correct option is C


Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

$36$ factorized into two factors in such a way that sum of factors is minimum, then the factors are

  1. $2, 18$
  2. $9, 4$
  3. $3, 12$

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

$36 = 1 \times 36$
     $= 2 \times 18$
     $= 3 \times 12$
     $= 4 \times 9$
     $= 6 \times 6$

1 + 36 = 37,  2 + 18 = 20,  3 + 12 = 15,  4 + 9 = 13,  6 + 6 = 12

Here, $12 < 13 < 15 < 20 < 37$


$\therefore \left( {6,6} \right)$ 

None of These