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 algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If (x+1) is a factor of $\displaystyle x^{3}+11x^{2}+15x+a$ then the value of 'a' is

  1. 2

  2. 3

  3. 5

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

$x+1\div x^{3}+11x^{2}+15x+a\setminus x^{2}+10x+5$

                $x^{3}+x^{2}$ ( subtracted)
               ..............................................................
                               $10x^{2}+15x+a$
                               $10x^{2}+10x$ ( subtracted)
                            ....................................................................
                                          $5x+a$
                                          $5x+5$  ( subtracted)
.....................................................................................................
                                             a-5
If x+1 is the factor of $x^{3}+11x^{2}+15x+a$
Then a-5=0 or a=5

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

________ is a method of writing numbers as the product of their factors or divisors.

  1. Polynomial

  2. Factorisation

  3. Division algorithm

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

Factorisation is a method of writing numbers as the product of their factors or divisors.
Example: $4x^2+2x$ is a factor $2x(2x+1)$
By multiplying the factor we get the original number.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

The value of  $k$  for which  $x - 1$  is a factor of the polynomial  $4 x ^ { 3 } + 3 x ^ { 2 } - 4 x + k$  is

  1. $3$
  2. $0$
  3. $1$
  4. $-3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$x-1$ is a factor of $4{x^2} + 3{x^2} - 4x + k$


put $x=1$


$4{x^2} + 3{x^2} - 4x + k=0$

$ \Rightarrow 4{\left( 1 \right)^2} + 3{\left( 1 \right)^2} - 4\left( 1 \right) + k = 0$

$ \Rightarrow 4 + 3 - 4 + k = 0$

$ \Rightarrow k =  - 3$

Hence,
option $(D)$ is correct answer.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Factorise : $6xy^2 + 4x^2y$ 

  1. $2xy(3x+y)$
  2. $xy(3x+2y)$
  3. $2xy(2x+3y)$

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

The common factor between $6xy^2$ and $4x^2y$ is $2xy$ that is the HCF of $6xy^2$ and $4x^2y$ is $2xy$. 

Therefore, we take $2xy$ as a common factor in the expression $6xy^2+4x^2y$ as shown below:
$6xy^2+4x^2y=2xy(2x+3y)$
Hence, the factors of $6xy^2+4x^2y$ are $2xy$ and $(2x+3y)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Which of the following is an example of factorisation?

  1. $x^2+2x=x(x+2)$
  2. $x^2+2x=x(x+1)$
  3. $x^2+2x=x(x+3)$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The common factor between $x^2$ and $2x$ is $x$ that is the HCF of $x^2$ and $2x$ is $x$. 


Therefore, we take $x$ as a common factor in the expression $x^2+2x$ as shown below:


$x^2+2x=x(x+2)$


Hence, the factorization of $x^2+2x$ is $x(x+2)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Factorise : $5mn+15mnp$

  1. $5mn(1 + 3p)$
  2. $3mn(1 + 5p)$
  3. $5mn(1 - 3p)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The common factor between $5mn$ and $15mnp$ is $5mn$ that is the HCF of $5mn$ and $15mnp$ is $5mn$. 


Therefore, we take $5mn$ as a common factor in the expression $5mn+15mnp$ as shown below:


$5mn+15mnp=5mn(1+3p)$


Hence, the factorization of $5mn+15mnp$ is $5mn(1+3p)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If $f(x)$ and $g(x)$ are two polynomials with integral coefficients which vanish at $x = \dfrac {1}{2}$, then what is the factor of HCF of $f(x)$ and $g(x)$?

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

Given, $x = \dfrac {1}{2}$

$ \Rightarrow (2x - 1) = 0$
Therefore, $ (2x - 1)$ is satisfying both $f(x)$ and $g(x)$, so $(2x - 1)$ is the factor of H.C.F. of $f(x)$ and $g(x)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Consider the following statements :
1. $x - 2$ is a factor of $x^{3} - 3x^{2} + 4x - 4$
2. $x + 1$ is a factor of $2x^{3} + 4x + 6$
3. $x - 1$ is a factor of $x^{6} - x^{5} + x^{4} - x^{3} + x^{2} - x + 1$
Of these statements

  1. 1 and 2 are correct

  2. 1, 2 and 3 are correct

  3. 2 and 3 are correct

  4. 1 and 3 are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
  1. Remainder $=2^3-3\times 2^2+4\times 2-4$
    $= 8-12+8-4 = 0$
    Hence $x-2$ is a factor.
    2. Remainder$= 2(-1)^3+4(-1)+6$
    $= -2-4+6 = 0$
    Hence $x + 1$ is a factor.
    3. Ramainder $= 1^6-1^5+1^5-1^3+1^2-1+1 = 1$
    Hence $x - 1$ is not a factor.
    $\therefore$ Statements 1 and 2 are correct.
Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If $4x^{4} -12x^{3}+x^{^{2}}+3ax-b$ is divided by $x^{2}-1$ then a = ____, and b=___

  1. $5, 4$
  2. $4,9$
  3. $4,5$
  4. $1, -1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ x }^{ 2 }-1=0\quad \Rightarrow x=\pm 1$

Given a polynomial $P(x)={ 4x }^{ 4 }-12{ x }^{ 3 }+{ x }^{ 2 }+3ax-b$
Using remainder theorem,as $P(x)$ is completely divisible by ${ x }^{ 2 }-1$
$\therefore P(\pm 1)=0\ \therefore P(1)=0\ \Rightarrow 4-12+1+3a-b=0\ \Rightarrow 3a-b=7\ \therefore P(-1)=0\ \Rightarrow 4+12+1-3a-b=0\ \Rightarrow 3a+b=17$
By soling both we get
$a=4$   ;$b=5$

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

$7+3x$ is a factor of $3x^3+7x$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $7 + 3x$ is a factor of $p\left( x \right) = 3{x^3} + 7x$, then, $p\left( {\frac{{ - 7}}{3}} \right) = 0$.

Compute $p\left( {\frac{{ - 7}}{3}} \right)$ in the given polynomial.

$p\left( {\frac{{ - 7}}{3}} \right) = 3{\left( { - \frac{7}{3}} \right)^3} + 7\left( {\frac{{ - 7}}{3}} \right)$

$ =  - \frac{{343}}{9} - \frac{{49}}{3}$

$ = \frac{{ - 343 - 147}}{9}$

$ =  - \frac{{490}}{9}$

The value of $p\left( {\frac{{ - 7}}{3}} \right)$is not equal to 0.

The given statement is false.

Multiple choice maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Find the common factors of the given terms:

$6 abc, 24ab^2, 12 a^2b$

  1. $6a^2b$
  2. $6ab^2$
  3. $6ab$
  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$6 abc$, $24ab^2$, $12 a^2b$
The factors of $6abc=2\times 3\times a\times b\times c$
The factors of $24ab^2=2\times 2\times 2\times 3\times a\times b\times b$
The factors of $12 a^2b=2\times 2\times 3\times a\times a\times b$
The common factors are $2\times 3\times a\times b=6ab$