Mathematics · Quantitative Aptitude

Algebraic Polynomials

138 Questions

Algebraic polynomials involve factoring expressions, applying the remainder theorem, and finding variable roots. These topics form a major part of the quantitative aptitude syllabus. Regular practice ensures accuracy in solving complex algebraic equations.

Factoring polynomialsFactor theoremRemainder theoremPolynomial rootsCubic polynomials

Algebraic Polynomials Questions

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

Consider the polynomial $\dfrac{x^{3}+2x+1}{5}-\dfrac{7}{2}x^{2}-x^{6}$. 

The constant term is:

  1. $\dfrac{1}{7}$
  2. $\dfrac{1}{5}$
  3. $\dfrac{1}{2}$
  4. $\dfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\cfrac { { x^{ 3 }+2x+1 } }{ 5 } -\cfrac { 7 }{ 2 } x^ 2-x^ 6$
$=\cfrac { { x^{ 3 }+2x+1 } }{ 5 } -\cfrac { 7 }{ 2 } x^ 2-x^ 6$
$=-x^ 6+\cfrac { x^{ 3 } }{ 5 } -\cfrac { 7x^ 2 }{ 2 } +\cfrac { 2x }{ 5 } +\cfrac { 1 }{ 5 }$
So, the constant term= $\cfrac { 1 }{ 5 }$

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

Which of the following expressions is a polynomial in one variable?

  1. $x+\dfrac {2}{3}+3$
  2. $3\sqrt {x}+\dfrac {2}{\sqrt {x}}+5$
  3. $\sqrt {2x^{2}}-\sqrt {3x}+6$
  4. $x^{10}+y^{5}+8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A polynomial in one variable must have non-negative integer exponents for the variable. Option A (x + 2/3 + 3) is a polynomial in x, whereas the others involve square roots of variables or multiple variables.

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

If $P(x)$ and $Q(x)$ are two polynomial such that $f(x)=P(x^3)+Q(x^3)$ is divisible by $x^2+x+1$, then?

  1. $P(x)$ is divisible by $(z-1)$ by $Q(x)$ is not divisible by $(x-1)$
  2. $Q(x)$ is divisible by $(x-1)$ but $P(x)$ is not divisible by $(x-1)$
  3. Both $P(x)$ and $Q(x)$ are divisible by $(x-1)$
  4. $f(x)$ is divisible by $(x-1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For f(x) to be divisible by x^2+x+1, it must vanish at the roots of x^2+x+1=0, which are omega and omega^2. Substituting x=omega into P(x^3)+Q(x^3) gives P(1)+Q(1)=0, implying P(1)=-Q(1). This condition is satisfied if both P(x) and Q(x) contain the factor (x-1).

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

What is the degree of the remainder atmost, when a fourth degree polynomial is divided by a quadratic polynomial?

  1. $2$
  2. $0$
  3. $4$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Here $f(x)$ represent dividend and $g(x)$ represent division

$g(x)=$ quadratic polynomial $=ax^2+bx+c$

$\therefore deg(g(x))=2$, $deg(f(x))=4$

quotient $q(x)$ is of degree $=2$ $(=4-2)$

Remainder $R(x)=$ degree $1$ or less than $1$.
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

Can $(x - 1)$ be the remainder on division of a polynomial $p(x)$ by $2x + 3$?

  1. Yes

  2. No

  3. Cannot be justified

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

Division algorithm stated that a polynomial $f(x)$ can written as
   $f(x) = g(x)q + r$     where $q$ and $r$ are unique integer and $0 <= r < g(x)$.
Here,
$g(x)=2x+3$ and $r(x)=x-1$
The power of the remainder is always less than the power of the divisor
Here, the degree of remainder is $1$ and the degree of divisor is $1$, which is not possible. Thus, $(x-1)$ cannot be the remainder of $p(x)$ when divided by $(2x+3)$

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

If on dividing a non-zero polynomial $p(x)$ by a polynomial $g (x)$, the remainder is zero, what is the relation between the degrees of $p(x)$ and $g (x)$?

  1. degree of $g (x) \ge$ degree of $p(x)$
  2. degree of $g(x) \le$ degree of $p(x)$
  3. degree of $g (x) =$ degree of $p(x)$

  4. Can't say
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

deg $p(x)=$ deg $g(x)+r(c)$

Then, deg $p(x) \ge$ deg $g(x)$

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

If the polynomial $x^3-x^2+x-1$ is divided by $x-1$, then the quotient is :

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

Divide $x^3-x^2+x-1$ by   $x-1$


         $x-1$ $\overline{)x^3-x^2+x-1(}$  $x^2+1$

                $-(x^3-x^2)$
                   $\overline{\quad\quad\quad\quad+x-1}$
                                   $-(x-1)$
                                   $\overline{\quad\quad\quad0}$

Hence, $B$ is correct.

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

When the polynomial  ${x^4} + {x^2} + 1$   is divided by $(x + 1)({x^2} - x + 1)$ then the remainder is $ax + b$ , then  $a + b$ is equal to 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{{x}^{4}+{x}^{2}+1}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

$=\dfrac{{x}^{4}+2{x}^{2}+1-{x}^{2}}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

$=\dfrac{{\left({x}^{2}+1\right)}^{2}-{x}^{2}}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

$=\dfrac{\left({x}^{2}-x+1\right)\left({x}^{2}+x+1\right)}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

$=\dfrac{\left({x}^{2}+x+1\right)}{\left(x+1\right)}$

$=\dfrac{x\left(x+1\right)+1}{\left(x+1\right)}$

$=x+\dfrac{1}{x+1}$

Remainder$=1$ is of the form $ax+b$

$\Rightarrow\,a=0,\,b=1$

$\therefore\,a+b=0+1=1$