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 business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

If $f(x)$ is a polynomial function satisfying the condition $f(x) \times f\left(\dfrac{1}{x}\right)=f(x)+f\left(\dfrac{1}{x}\right)$ and $f(2)=9$ then

  1. $2f(4) =3 f(6)$
  2. $14f(1) = f(3)$
  3. $ 9f(3) = 2f(5)$
  4. $f(10) = f(11)$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

The polynomial which satisfies $f(x)f(1/x)=f(x)+f(1/x)$ is $ \pm x^n+1$ (standard result)
Given that $f(2) = 9 \ \Rightarrow \pm 2^n + 1 = 9 \ \Rightarrow 2^n = 8 $
(-ve sign not possible here)
$ \Rightarrow n=3$
Hence the function is $ f(x)=x^3+1$
$ \Rightarrow f(1) = 2, \; f(3)=28 , \; f(5)=126$
$ f(4) = 65, \; f(6) = 217$
Using these, we see only option B and C are correct. 

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

If $g(x)$ is a polynomial satisfying $g(x) g(y) = g(x) + g(y) + g(xy) - 2$ for all real $x$ and $y$ and $g(2) = 5$ then $g(3)$ is equal  to -

  1. $10$
  2. $24$
  3. $21$

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

$g(x) g(y) = g (x) + g (y) + g (xy) - 2$


Substitute $x = 2$ & $y = 1$

$g (2) g (1) = g (2) + g (1) + g (2) - 2$

$\Rightarrow 4g (1) = 8 $

$\Rightarrow g (1) = 2$

$g (x) g(y) = g (x) + g (y) + g(xy)-2$, now substitute $y\, =\, \displaystyle \frac{1}{x}$

Now $g (x)\, g \left (\displaystyle \frac{1}{x} \right )\, =\, g (x)\, +\, g \left (\displaystyle \frac{1}{x} \right )$

$g(x) = 1\, \pm\, x^n$

$\therefore\, 5\, =\, 1\, \pm\, 2^n\, (\because\, g (2)\, =\, 5)$

$2^n=4$

So, $n = 2$

Now $g (3) = 1$ + $3^2=10$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $f(x)$ be a non-zero polynomial of degree $4$. Extreme points of $f(x)$ are $0, -1, 1$. If $f(k)=f(0)$ then?

  1. k has one rational & two irrational roots

  2. k has four rational roots

  3. k has four irrational roots

  4. k has three irrational roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $f'(x)=\lambda x(x^2-1)\Rightarrow f(x)=\lambda\left(\dfrac{x^4}{4}-\dfrac{x^2}{2}\right)+C$
Now $f(0)=f(k)\Rightarrow \dfrac{k^4}{4}-\dfrac{k^2}{2}=0\Rightarrow k=0$ or $\pm \sqrt{2}$
Hence $(1)$.

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

Is the following quadratic polynomial reducible or irreducible?
$f(x) = -2x^2-2x-1$

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

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

To check whether the given quadratic polynomial is reducible or irreducible, we need to calculate the discriminant

Calculate the discriminant for the equation, $-2x^2-2x-1=0$

$D=b^{2}-4ac=(-2)^2-4(-2)(-1)=-4<0$
Quadratic equation is irreducible if $D<0$
$\therefore$ The quadratic polynomial is irreducible.

Correct option is C

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

When multiplicity of a polynomial exist?

  1. when a factor appears in conjugate

  2. when a factor appears more than once

  3. when a factor appears more than twice

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

"multiplicity of a polynomial exist when a factor of a polynomial repeats more than once"

For eg. In the polynomial $(x -2)^3(x-3)^2(x-1 )$ 
The root $2$ have multiplicity $3$, the root  $3$ have $2$, and the root $1$ have $1$
Hence, B is correct.

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

Find multiplicity of the polynomial
$f(x) = (x-1)^2(2x+5)^3(x^2+1)^2(x+\pi^2)^4$

  1. $4$
  2. $2$
  3. $8$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of the given function are $1,-\frac{5}{2} ,-i,i,-\pi^{2}$

The multiplicity of$x=1$ is $2$
The multiplicity of $x=-\frac{5}{2}$ is $3$
The multiplicity of $x=i $ is $2$
The multiplicity of $x=-i$ is $2$
The multiplicity of $x=-\pi^{2}$ is $4$
Therefore the multiplicity of polynomial is $4$
Therefor option $A$ is correct

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

List the multiplicities of the zeroes of the polynomial $P(x)=x^2-14x+49$

  1. $x=5$ is a zero of multiplicity $3$.
  2. $x=7$ is a zero of multiplicity $2$.
  3. $x=6$ is a zero of multiplicity $1$.

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

Given, $x^2-14x+49$

$\Rightarrow x^2-7x-7x+49$
$\Rightarrow x(x-7)-7(x-7)$
$\Rightarrow (x-7)^2$
So, $x=7$ is a zero of multiplicity $2$.

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

Is the following quadratic polynomial reducible or irreducible?
$f(x) = x^2 - \sqrt2$

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

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

Let $\sqrt 2=a^2\Rightarrow a^2=2^{1/2}\Rightarrow a=2^{1/4}$

So $f(x)=x^2-a^2=(x+a)(x-a)=(x+2^{1/4})(x-2^{1/4})$, using $a^2-b^2=(a+b)(a-b)$
So given quadratic is reducible to two real roots 

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If $f(x),g(x)$ and $h(x)$ are three polynomials of degree $2$ and $\Delta(x) \left| \begin{matrix} f\left( x \right)  \ f'\left( x \right)  \ f"\left( x \right)  \end{matrix}\begin{matrix} g\left( x \right)  \ g'\left( x \right)  \ g"\left( x \right)  \end{matrix}\begin{matrix} h\left( x \right)  \ h'\left( x \right)  \ g"\left( x \right)  \end{matrix} \right|$ then polynomial of degree (whenever defined)

  1. $2$
  2. $3$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\triangle \left( x \right) =\begin{vmatrix} f\left( x \right)  & g\left( x \right)  & h\left( x \right)  \\ f^{ ' }\left( x \right)  & g^{ ' }\left( x \right)  & h^{ ' }\left( x \right)  \\ f^{ '' }\left( x \right)  & g^{ '' }\left( x \right)  & h^{ '' }\left( x \right)  \end{vmatrix}$
$\Rightarrow f\left( x \right) \left[ g^{ ' }\left( x \right) .h^{ '' }\left( x \right) -h^{ ' }\left( x \right) g^{ '' }\left( x \right)  \right] -g\left( x \right) \left[ f^{ ' }\left( x \right) h^{ '' }\left( x \right) -h^{ ' }\left( x \right) f^{ '' }\left( x \right)  \right] +h\left( x \right) \left[ f^{ ' }\left( x \right) g^{ '' }\left( x \right) -g^{ ' }\left( x \right) f^{ '' }\left( x \right)  \right] $
$\because f\left( x \right) ,g\left( x \right) ,h\left( x \right) $ are polynomial of degree $2$
$\therefore f^{ ' }\left( x \right) ,g^{ ' }\left( x \right) ,h^{ ' }\left( x \right) $ are of degree $1$
$\therefore f^{ '' }\left( x \right) ,g^{ '' }\left( x \right) ,h^{ '' }\left( x \right) $ are of degree $0$
$\therefore \triangle \left( x \right) $ is polynomial of degree $3$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

The sum and the product of the zeroes of a quadratic polynomial are $ \dfrac{-1}{2} $ and $ \dfrac{1}{2}$ respectively, then the polynomial is :

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

Given: Sum of zeroes $=-\dfrac 12$ and product of zeroes $=\dfrac 12$

We know,
$x^2-(\text{sum of zeroes})x+(\text{product of zeroes})=0$
$\Rightarrow x^2-\left(-\dfrac 12\right)x+\dfrac 12=0$
$\Rightarrow 2x^2+x+1=0$
is the required polynomial.