Mathematics · Quantitative Aptitude

Algebraic Expressions and Polynomials

275 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 brackets order operations and algebra using brackets in algebraic expressions order of operations

Solve the expression using BODMAS rule: $3x(x-2)+x(x^2\times 2x)-12x$

  1. $2x^5-3x^2-18x$
  2. $2x^5+3x^2-18x$
  3. $2x^5+3x^2+18x$
  4. $-2x^5-3x^2-18x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$3x(x-2)+x(x^2\times 2x)-12x$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $3x(x-2)+x(x^2\times 2x)-12x$
$=$ $3x^2-6x+x^3\times2x^2-12x$
$=$ $3x^2-6x+2x^5-12x$
$=$ $2x^5+3x^2-18x$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

Solve the following equations:
$x^{2} + 2xy + 3xz = 50$,
$2y^{2} + 3yz + yx = 10$,
$3z^{2} + zx + 2zy = 10$.

  1. $x=\pm 4; y=\pm 2; z=\pm 2$
  2. $x=\pm -4; y=\pm -2; z=\pm 2$
  3. $x = \pm 5; y = \pm 1; z = \pm 1$

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

Given equations are ${ x }^{ 2 }+2xy+3xz=50$

$\Rightarrow  x(x+2y+3z)=50$    ........(i),
$ 2{ y }^{ 2 }+3yz+yx=10$
$\Rightarrow  y(2y+3z+x)=10$    ........(ii)
and $ 3{ z }^{ 2 }+zx+2zy=10$
$\Rightarrow  z(3z+x+2y)=10$    ..........(iii)
Dividing (i) by (ii), we get

$\dfrac { x }{ y } =\dfrac { 50 }{ 10 } =5\\ \Rightarrow x=5y$    ......(a)
Dividing (ii) by (iii), we get
$\dfrac { y }{ z } =\dfrac { 10 }{ 10 } =1\\ \Rightarrow z=y$    ..........(b)
Substituting (a) and (b) in (ii)
$y(2y+3y+5y)=10\\ \Rightarrow 10{ y }^{ 2 }=10\\ \Rightarrow y=\pm 1$
From (a), we have
$x=\pm 5$
From (b), we have
$z=\pm 1$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

Solve the following equations:
$x + 2y - z = 11$,
$x^{2} - 4y^{2} + z^{2} = 37$,
$xz = 24$.

  1. $x=2, -5; y=2; z=2,-4$
  2. $x=8,-3; y=3; z=-3,-8$
  3. $x=-3, 5; y=4; z=2,5$
  4. $x=2,4; y=3; z=3, -5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ x }^{ 2 }-4{ y }^{ 2 }+{ z }^{ 2 }=37$    ......(i)

$xz=24$    .......(ii)
$ x+2y-z=11$    .....(iii)
$\Rightarrow  x-z=11-2y$
On squaring both sides, we have

${ x }^{ 2 }+{ z }^{ 2 }-2xz=121+4{ y }^{ 2 }-44y\\ \Rightarrow { x }^{ 2 }+{ z }^{ 2 }-4{ y }^{ 2 }-2xz=121-44y\\ \Rightarrow 37-2(24)=121-44y\\ \Rightarrow -44y=-132\\ \Rightarrow y=3$
Substituting $y$ in (iii), 
$x+6-z=11\\ \Rightarrow x-z=5$
From (ii), $z=\dfrac { 24 }{ x } $
Thus $x-\dfrac { 24 }{ x } =5$
$ \Rightarrow { x }^{ 2 }-24=5x\\ \Rightarrow { x }^{ 2 }-5x-24=0\\ \Rightarrow { x }^{ 2 }-3x+8x-24=0\\ \Rightarrow x(x-3)+8(x-3)=0\\ \Rightarrow (x+8)(x-3)=0\\ \Rightarrow x=-8,3$
Putting in $ z=\dfrac { 24 }{ x } $
Thus $ z=-3,8$
So, the values of $x$ are $-8,3$, values of $z$ are $-3,8$ and value of $y$ is $3$.

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 reciprocal equations theory of equations maths

The number of solutions $(x, y, z)$ to the system of equations $ x + 2y + 4z = 9, 4yz + 2xz + xy = 13, xyz = 3 $ such that at least two of $ x, y, z$ are integers is

  1. $3$
  2. $5$
  3. $6$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the roots of the system equation are:
$\alpha =x,\beta =2y,\gamma =4z$
$\alpha +\beta +\gamma =x+2y+4z=9$
$\alpha \beta +\beta \gamma +\gamma \alpha =2xy+8yz+yzx$
$=2(4yz+2xz+xy)\Rightarrow 26$
$\alpha \beta \gamma =8xyz\Rightarrow 24$
Thus,our polynomial should be:
$P^{3}-9P+26P-24=0$
$(P-2)(P-3)(P-4)=0$
since our roots are :
$\alpha =x,\beta =2y$ and $\gamma =4z$
$(x,2y,4z)=(2,3,4)$ or its permutations,or 6 combination.
However,note that one case if,
$x=4,2y=3$ and $4z=2$
$(x,y,z)=(4,\dfrac{3}{2},\dfrac{1}{2})$
which two of the roots are not an integer :Excluding of this case ,we have five solutions.
Multiple choice maths part number dividing fractions division of a fractions division of a fraction

For $a = 4$, it is known that the value of the fraction $\dfrac{(a+2)x + a^2-1}{ax-2a +18}$ is independent of $x$. The other values of a for which this is the case, belong to the interval 

  1. $[-\infty, -2]$
  2. $[-2, 0]$
  3. $[0, 2]$
  4. $[2, 4]$
  5. $[4, +\infty]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\cfrac { \left( a+2 \right) x+{ a }^{ 2 }-1 }{ ax-2a+18 } $ is independent of x.

$\cfrac { a+2 }{ a } =\cfrac { { a }^{ 2 }-1 }{ 18-2a } \ 18a+36-2{ a }^{ 2 }-4a={ a }^{ 3 }-a\ { a }^{ 3 }+2{ a }^{ 2 }-15a-36=0\ { a }^{ 3 }-4{ a }^{ 2 }+6{ a }^{ 2 }-24a+9a-36=0\ (a-4)({ a }^{ 2 }+6a+9)=0\ \therefore a=-3,-3$ 
Other values of a belongs to $(-\infty ,-2]$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Given, $y=3$, $y=ax^2+b$
In the system of equations above, $a$ and $b$ are constants. For which of the following values of $a$ and $b$ does the system of equations have exactly two real solutions?

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

On substituting value of $y=3$ in second equation, we get
$ax^2+b=3$
$\Rightarrow ax^2+b-3=0$
$\Rightarrow ax^2=3-b$
$\Rightarrow x^2=\frac{3-b}a$
Since $x^2$ is positive quantity, therefore just $a=2$ and $b=2$ satisfies this.
Hence, option A is correct.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The system of equations:

$\displaystyle y={ x }^{ 2 }-2x$
$\displaystyle y=2x-1$ has two solutions for ($x,y$). 
Determine the greater value of $x$.

  1. $\displaystyle 2-\sqrt { 3 } $
  2. $\displaystyle \sqrt { 3 } $
  3. $\displaystyle 2+2\sqrt { 3 } $
  4. $\displaystyle 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given, $y=x^{2}-2x$
$y=2x-1$
Then $x^{2}-2x=2x-1 $
$\Rightarrow x^{2}-2x-2x+1=0$
$\Rightarrow x^{2}-4x+1=0$
Manipulate this equation $ax^{2}+bx+c=0$
We know $x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\therefore x=\dfrac{-(-4)\pm \sqrt{(-4)^{2}-4(1)(1)}}{2(1)}$
$\Rightarrow x= \dfrac{4\pm \sqrt{16-4}}{2}$
$\Rightarrow x=\dfrac{4\pm \sqrt{12}}{2}$
$\Rightarrow x=\dfrac{4\pm 2\sqrt{3}}{2}$
The greater of the two possible values for $x$ is $x=2\pm 2\sqrt{3}$. 
Therefore, the correct answer is (C).
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 $a^{2}-b^{2} = (a-b)(a+b)$

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

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