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 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 zeroes of a quadratic polynomial $p(x)$ are $-7$ and $-10$ respectively. Then $p(x)$ is :

  1. $x^{2}-7x-10$
  2. $x^{2}-7x+10$
  3. $x^{2}+7x-10$
  4. $x^{2}+7x+10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given: Sum of zeroes $=-7$ and product of zeroes $=-10$
We know that
$p(x)=x^2-(\text{sum of zeroes})x+(\text{product of zeroes})$
$\Rightarrow p(x)=x^2-(-7)x+(-10)$
$\Rightarrow p(x)=x^2+7x-10$
is the required polynomial.
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

A quadratic polynomial $p(x)$ with $3$ and $\dfrac{-2}{5}$  as the sum and product of zeroes, respectively is $10x^2+30x-4$

  1. True

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

$10x^2+30x-4=0$

$\alpha+\beta=-\cfrac{30}{10}=-3$
$\alpha\beta=\cfrac{-4}{10}=\cfrac{-2}{5}$
So, a quadratic polynomial $p(x)$ with $3$ and $\cfrac{-2}{5}$ as the sum and product of zeroes, respectively is $10x^2+30x-4$ is false.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

A polynomial of 6th degree $f(x)$ satisfies $f(x)=f(2-x),:\forall:x\epsilon R$, if $f(x)=0$ has 4 distinct and two equal roots, then sum of the roots of $f(x)=0$ is:

  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the functional relation replace $x$ with $x+1$.

We have,
$f(1+x)=f(1-x)$
This shows that the function is symmetric about $x=1$.
There is one and only one double root. If the double root exists at any value $x _0$ other than at $x=1$, then a double root will also exist at a value of $2-x _0$.
Hence, the double root exists at $x=1$ 
Say two other roots are $\alpha$ and $\beta$
$f(\alpha)=f(2-\alpha)=0$
$\therefore 2-\alpha$ is also a root.
And similarly, $2-\beta$ is also a root.
$\therefore$ the roots are $1, 1, \alpha, \beta, 2-\alpha, 2-\beta$
Hence, sum of the roots is $6$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The polynomial $\displaystyle (ax^{2}+bx+c)(ax^{2}-dx-c),ac\neq 0,$ has

  1. four real zeros

  2. at least two real zeros

  3. at most two real zeros

  4. no real zeros
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 
$ax^{2}+bx+c=0$
$b^{2}-4ac\geq 0$ for real roots ...(i)
and for 
$ax^{2}-dx-c=0$
$d^{2}+4ac\geq 0$ for real roots ...(ii)
Now, 
$ac\neq 0$
Hence, 
Case I
If $ac>0$
Hence, 
$ax^{2}-dx-x=0$ has positive roots.
Case II
If $ac<0$
Then,
$ax^{2}+bx+c=0$
has Real roots.
Hence, the above polynomial has atleast two real roots.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $\alpha$ and $\beta$ are the zeros of polynomial $x^{2}-ax+b$, then the value of $\alpha^{2}\left(\dfrac {\alpha^{2}}{\beta}-\beta\right)+\beta^{2}\left(\dfrac {\beta^{2}}{\alpha}-\alpha\right)$ is

  1. $\dfrac {a(a^{2}-4b)(a^{2}-b)}{b}$
  2. $\dfrac {b(a^{2}-4b)(a^{2}-b)}{a}$
  3. $\dfrac {b^{2}(a^{2}-4b)(a^{2}-b)}{a}$

  4. None

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have Sum of the roots$=\alpha+\beta=a$
Product of the roots$=\alpha\beta=b$

${\alpha}^{2}\left(\dfrac{{\alpha}^{2}}{\beta}-\beta\right)+{\beta}^{2}\left(\dfrac{{\beta}^{2}}{\alpha}-\alpha\right)$
$=\dfrac{{\alpha}^{2}}{\beta}\left({\alpha}^{2}-{\beta}^{2}\right)+\dfrac{{\beta}^{2}}{\alpha}\left({\beta}^{2}-{\alpha}^{2}\right)$
$=\dfrac{{\alpha}^{2}}{\beta}\left({\alpha}^{2}-{\beta}^{2}\right)-\dfrac{{\beta}^{2}}{\alpha}\left({\alpha}^{2}-{\beta}^{2}\right)$
$=\left({\alpha}^{2}-{\beta}^{2}\right)\left(\dfrac{{\alpha}^{2}}{\beta}-\dfrac{{\beta}^{2}}{\alpha}\right)$
$=\dfrac{\left({\alpha}^{2}-{\beta}^{2}\right)}{\alpha\beta}\left({\alpha}^{3}-{\beta}^{3}\right)$
$=\dfrac{\left(\alpha-\beta\right)\left(\alpha+\beta\right)}{\alpha\beta}\left(\alpha-\beta\right)\left({\alpha}^{2}+{\beta}^{2}+\alpha\beta\right)$
$=\dfrac{{\left(\alpha-\beta\right)}^{2}\left(\alpha+\beta\right)}{\alpha\beta}\left({\alpha}^{2}+{\beta}^{2}+\alpha\beta\right)$

We know that ${\alpha}^{2}+{\beta}^{2}={\left(\alpha+\beta\right)}^{2}-2\alpha\beta$ and 
${\left(\alpha-\beta\right)}^{2}={\left(\alpha+\beta\right)}^{2}-4\alpha\beta$

Using $\alpha+\beta=a$ and $\alpha\beta=b$ we have
${\left(\alpha-\beta\right)}^{2}={\left(\alpha-\beta\right)}^{2}={a}^{2}-4b$
And ${\alpha}^{2}+{\beta}^{2}+\alpha\beta={\left(\alpha+\beta\right)}^{2}-2\alpha\beta+\alpha\beta$
$={\left(\alpha+\beta\right)}^{2}-\alpha\beta={a}^{2}-b$
$=\dfrac{a\left({a}^{2}-4b\right)\left({a}^{2}-b\right)}{b}$



Multiple choice maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

The expression $(x + 1)(x + 2)(x + 3)(x + 4) + 1$ is a 

  1. perfect square

  2. cube

  3. quartic polynomial

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

Solving

$(x+1)(x+2)(x+3)(x+4)+1$
$Multiplying\ first\ bracket\ with\ last\ and\ second\ to\ the\ third\ one$
$(x^2+5x+4)(x^2+5x+6)+1$
$Replacing\ x^2+5x+4\ by\ 'B'$
$(B)(B+2)+1$
$B^2+2B+1$
$(B+1)^2=(x^2+5x+5)^2$
Hence $L.H.S.$ is the $Perfect\ Square$ of $(x^2+5x+5)$



Multiple choice business economics and quantitative methods correlation analysis aspects of correlation scatter graphs and correlation linear regression

If two polynomial of the degree 7 and 4 respectively are multiplied, find the degree of the resultant polynomial

  1. 11

  2. -11

  3. 12

  4. 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When multiplying two polynomials, the degree of the resulting polynomial is the sum of the degrees of the original polynomials. Adding 7 and 4 results in 11.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

What is the HCF of the polynomials $x^{4} - 3x + 2, x^{3} - 3x^{2} + 3x - 1$ and $x^{4} - 1$?

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

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

$x^{4} - 3x + 2 = x^{4} - x^{3} + x^{3} - x^{2} + x^{2} - x - 2x + 2$
$= x^{3} (x - 1) + x^{2} (x - 1) + x(x - 1) - 2(x - 1)$
$= (x - 1) [x^{3} + x^{2} + x - 2]$
$x^{3} - 3x^{2} + 3x - 1 = (x - 1)^{3}$
$x^{4} - 1 = (x - 1) (x + 1) (x^{2} + 1)$
$HCF = x - 1$

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If the HCF of the polynomials $f(x)$ and $g(x)$ is $4x - 6$, then $f(x)$ and $g(x)$ could be :

  1. $2, 2x - 3$
  2. $8x - 12, 2$
  3. $\displaystyle 2\left ( 2x-3 \right )^{2},4\left ( 2x-3 \right )$
  4. $\displaystyle 2\left ( 2x+3 \right ),4\left ( 2x+3 \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, HCF $ = 4x-6 = 2(2x-3) $

Since HCF needs to be a factor of both the polynomials, clearly only option C with polynomials $ 2({2x-3)}^{2} , 4(2x-3) $  have both factors $ 2 $ and $ (2x-3) $

Multiple choice biology classification of living organisms binomial nomenclature nomenclature of organisms naming of the organisms nomenclature in classification

Polynomial classification has recently begun to replace the binomial system

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Polynomial System of nomenclature was the older form of naming organisms before the advent of binomial system. In polynomial system, a generic name and specific name which consisted of many words were used. It was eventually found to be incompatible and hence was replaced by binomial system.
So, the correct answer is, 'False'.