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

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder. 
$p(x)=x^4-3x^2+4x+5$
$g(x)=x^2+1-x$

  1. $q(x)=x^2+x-3$ and $r(x)=-8$
  2. $q(x)=x^2-x+3$ and $r(x)=8$
  3. $q(x)=x^2+x-3$ and $r(x)=8$
  4. $q(x)=x^2-x-3$ and $r(x)=-8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2-x+1)\overline {x^4-3x^2+4x+5}$ ( $x^2+x-3$
                  $\underline {\underset {-}{}x^4\underset {-}{+}x^2              \underset{+}{-}x^3}$
                  $x^3-4x^2+4x+5$
                  $\underline {\underset {-}x^3\underset {+}{-}x^2\underset {-}{+}x}$
                  $-3x^2+3x+5$
                  $\underline {\underset {+}{-}3x^2\underset {-}{+}3x\underset {+}{-}3}$
                                  $8$
Hence, Quotient=$x^2+x-3$
Remainder=8.

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

Polynomials $p(x), g(x), q(x)$ and $r(x)$, which satisfy the division algorithm and deg $r(x)=0$, are

  1. $p(x)=x^2+x; g(x)=x+1$;
    $q(x)=5; r(x)=7$.
  2. $p(x)=2x^2+x+1; g(x)=x+3$;
    $q(x)=2x; r(x)=7x$.
  3. $p(x)=x^3+x+5; g(x)=x^2+1$;
    $q(x)=x; r(x)=5$.

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

option (A) and have deg $r(x)=0$,

in this question $p(x)=q(x)g(x)+r(x)$ is satisfied in only option (C).

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

On dividing $f(x)$ by a polynomial $x-1-x^2$, the quotient $q(x)$ and remainder $r(x)$ are $(x-2)$ and $3$ respectively. Then $f(x)$ is

  1. $f(x)=-3x^2-x+7$
  2. $f(x)=-x^3+x^2-x+7$
  3. $f(x)=3x^2-3x+5$
  4. $f(x)=-x^3+3x^2-3x+5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f(x)=q(x)g(x)+r(x)$

$\therefore f(x)= (x-2)(x-1-x^2)+3$

$\Rightarrow  f(x)= x(x-1-x^2)-2(x-1-x^2)+3$

$=x^2-x-x^3-2x+2+2x^2+3$

$=-x^3+3x^2-3x+5$

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

On dividing $x^3-3x^2+x+2$ by a polynomial $g(x)$, the quotient and remainder were $(x-2)$ and $(-2x+4)$, respectively. Find $g(x)$.

  1. $2x^2+2x-8$
  2. $x^2+2x-7$
  3. $x^2-x+1$
  4. $2x^2-x+2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
By Remainder theorem,
$p(x)=g(x)q(x)+r(x)$

We have, $p(x)=x^3-3x^2+x+2,q(x)=x-2\space and \space r(x)=-2x+4$

$\therefore x^3-3x^2+x+2=g(x)(x-2)+(-2x+4)$

$\Rightarrow x^3-3x^2+x+2+2x-4=g(x)(x-2)$

$\Rightarrow g(x)=\dfrac{x^3-3x^2+3x-2}{(x-2)}=\dfrac{x^3-2x^2-x^2+2x+x-2}{(x-2)}$
                                         
$=\dfrac{[x^2(x-2)-x(x-2)+1(x-2)]}{(x-2)}$

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

$=x^2-x+1$

$\therefore g(x)=x^2-x+1$
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

Polynomials $p(x), g(x), q(x)$ and $r(x)$, which satisfy the division algorithm and "deg $q(x) = $ deg $ r(x)$", are

  1. $p(x)=2x^2+x; g(x)=2x^2-4$;
    $q(x)=2x-7; r(x)=-x+2$
  2. $p(x)=x^2+x-3; g(x)=x^2+x-1$;
    $q(x)=7; r(x)=-5$
  3. $p(x)=x^2+x; g(x)=x^2-4$;
    $q(x)=2x-1; r(x)=-x-2$
  4. $p(x)=2x^2+2x+8; g(x)=x^2+x+9$;
    $q(x)=2; r(x)=-10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

according to division algorithm $p(x)=q(x)g(x)+r(x)$

degree of $q(x)$ is equal to $r(x)$ in all options.
only (D) option satisfies $p(x)=q(x)g(x)+r(x)$
$g(x)q(x)=2(x^2+x+9)=2x^2+2x+18=p(x)+10=p(x)-r(x)$ hence $p(x)=q(x)g(x)+r(x)$ in (D) satisfies division algorithm

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

Polynomials $p(x), g(x), q(x)$ and $r(x)$, which satisfy the division algorithm and "deg $p(x) = $ deg $q(x)$" are

  1. $p(x)=2x^2+2x+8, g(x)=4x+1$;
    $q(x)=x^2; r(x)=1$
  2. $p(x)=2x^2+2x+8, g(x)=5$;
    $q(x)=4; r(x)=4x-1$
  3. $p(x)=2x^2+2x+8, g(x)=2$;
    $q(x)=x^2+x+4; r(x)=0$
  4. $p(x)=x^2+x+3, g(x)=2x+3$;
    $q(x)=2x^2+x; r(x)=3x-2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

degree of $p(x)$ and $q(x)$ are equal in (A),(C),(D)

according to division algorithm, $p(x)=q(x)g(x)+r(x)$
in option (C), $g(x)q(x)=2(x^2+x+4)=2x^2+2x+8+0=g(x)q(x)+r(x)=p(x)$
hence option (C) is correct answer.

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

What should be added to $8x^4+14x^3-2x^2+7x-8$ so that the resulting polynomial is exactly divisible by $4x^2+3x-2$?

  1. $10-14x$
  2. $4x-10$
  3. $3x-5$
  4. $5-3x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$4x^2+3x-2)\overline {8x^4+14x^3-2x^2+7x-8}$ ( $2x^2+2x-1$
                          $\underline {\underset {-}{8}x^4\underset {-}{+}6x^3\underset {+}{-}4x^2}$
                                     $8x^3+2x^2+7x-8$
                                     $\underline {\underset {-}{8}x^3\underset {-}{+}6x^2\underset {+}{-}4x}$
                                             $-4x^2+11x-8$
                                             $\underline {\underset {+}{-}4x^2\underset {+}{-}3x\underset {-}{+}2}$
                                                          $14x-10$
We have to add $10-14x$ so that $8x^4+14x^3-2x^2+7x-8$ is completely divisible by $4x^2+3x-2$.

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

Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm. $x^3-3x+1, x^5-4x^3+x^2+3x+1$

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x^3-3x+1)\overline {x^5-4x^3+x^2+3x+1}$($x^2-1$
                         $\underset {-}{x^5}\underset {+}{-}3x^3\underset {-}{+}x^2$
                         $\overline {-x^3+3x+1}$
                         $\underline {\underset {+}{-}x^3\underset {-}{+}3x\underset {+}{-}1}$
                                              $2$
Since remainder is non-zero.
Therfore,$x^3-3x+1$ is not a factor of $x^5-4x^3+x^2+3x+1$

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 $f(x)=x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the remainder comes out to be $(x+a)$, then values of $k$ and $a$ are

  1. $k=-2$ & $a=4$
  2. $k=5$ & $a=-5$
  3. $k=-3$ & $a=-7$

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

$f(x)=$ is divided by another polynomial
$x^2-2x+k)\overline {x^4-6x^3+16x^2-25x+10}(x^2-4x+(8-k)$
                       $\underline {\underset {-}{x^4}\underset {+}{-2x^3}\underset{-}{+}kx^2}$
                       $-4x^3+(16-k)x^2-25x+10$
                       $\underline {\underset {-}{-4x^3}\underset {-}{+8x^2}                      \underset{+}{-}4kx}$
                       $(8-k)x^2+(4k-25)x+10$
                       $\underline {\underset {-}(8-k)x^2+\underset {-}(2k-16)x\underset{-}{+}(8k-k^2)}$
                       $(2k-9)x+(k^2-8k+10)$
But remainder is given $x+a$
$\therefore x+a=(2k-9)x+(k^2-8k+10)$
On equating coefficient, we get
$2k-9=1\Rightarrow k=5$
and $a=k^2-8k+10\Rightarrow a=25-40+10=-5$
Hence, $k=5,a=-5$

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

Find the value of $b$ for which the polynomial $2x^3+9x^2-x-b$ is exactly divisible by $2x+3$?

  1. $15$
  2. $-15$
  3. $10$
  4. $-10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $2x+3$ is a factor of the polynomial $p\left(x\right)=2x^3+9x^2-x-b$
Therefore, by Factor theorem $p\left(-\dfrac32\right)=0$
$\Rightarrow 2\left(-\dfrac32\right)^3+9\left(-\dfrac32\right)^2-\left(-\dfrac32\right)-b=0$

$\Rightarrow -\dfrac{27}4+\dfrac{81}4+\dfrac32-b=0$
$\Rightarrow \dfrac{-27+81+6}4-b=0\Rightarrow b=\dfrac{60}4=15$
$\therefore \space b=15$

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

A polynomial when divided by $\displaystyle \left ( x-6 \right )$ gives a quotient $\displaystyle x^{2}+2x-13$ and leaves a remainder $-8$. Then polynomial is

  1. $\displaystyle x^{3}+4x^{2}+25x-78$
  2. $\displaystyle x^{3}-4x^{2}-25x+70$
  3. $\displaystyle x^{3}-4x^{2}-25x-70$
  4. $\displaystyle x^{3}+4x^{2}-25x+78$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $P$ be the polynomial. If $P$ is divided by $(x-6)$ then it leaves a remainder $-8$ and gives a quotient $x^2+2x-13$. Therefore, 

$\cfrac { P }{ x-6 } ={ x }^{ 2 }+2x-13-\cfrac { 8 }{ x-6 } \\ \Rightarrow P=(x-6)({ x }^{ 2 }+2x-13)-\frac { 8(x-6) }{ x-6 } \\ \Rightarrow P={ x }^{ 3 }+2{ x }^{ 2 }-13x-6{ x }^{ 2 }-12x+78-8\\ \Rightarrow P={ x }^{ 3 }-4{ x }^{ 2 }-25x+70$

Hence, the polynomial is $x^3-4x^2-25x+70$.
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

If $\displaystyle f(x)=x^{4}-2x^{3}+3x^{2}-ax+b$ is a polynomial such that when it is divided by $( x - 1 )$ and $( x +1)$, the remainders are $5$ and $19 $ respectively, the remainder when $f(x)$ is divisible by $(x -2 ) $ is 

  1. $7$
  2. $8$
  3. $9$
  4. $10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When ${ x }^{ 4 }-{ 2x }^{ 3 }+3{ x }^{ 2 }-ax+b$ is divide by $ x-1,$ remainder is $5.$
So, substituting for $x$ is $1,$ in the above, we get

$5=1-2+3-a+b$ 
$\therefore-a+b=3 $----(1)

When ${ x }^{ 4 }-{ 2x }^{ 3 }+3{ x }^{ 2 }-ax+b$ is divide by $x+1$, remainder is $19.$
So, substituting for $x$ is $-1,$ in the above, we get 
$19=1+2+3+a+b$
$a+b=13 $----(2)
Solving (1) and (2), we get $a=5,b=8$
So polynomial becomes  ${ x }^{ 4 }-{ 2x }^{ 3 }+3{ x }^{ 2 }-5x+8$
The remainder when ${ x }^{ 4 }-{ 2x }^{ 3 }+3{ x }^{ 2 }-5x+8$ is divided by $x-2$ is by plugging in $x$ as $2$ in the given polynomial, we get 

$16-16+12-10+8=10$
so remainder is $10$
So, option D.

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

The remainders of polynomial f(x) when divided by x-1, x-2 are 2,3 then the remainder of f(x) when divided by (x-1) (x-2) is

  1. 2x-1

  2. x-1

  3. 2x+1

  4. x+1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
According to Remainder theorem

$f(x)=(x-1)(x-2) \theta (x)+\gamma (x)$

$\gamma (x)=ax+b$

$f(1)=a+b=2$     $\dots(1)$

$f(2)=2a+b=3$     $\dots(2)$

Subtract $(1)$ from $(2)$

$2a+b-a-b=3-2$

$\Rightarrow a=1$ substitute in $(1)$

$b=2-1=1$

$\therefore\ a=b=1$

So, $\gamma (x)=x+1$

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

If the remainders of the polynomial f(x) when divided by x+1 and x-1 are 3, 7 then the remainder of f(x) when divided by $(x^{2} -1 )$ is

  1. x + 4

  2. 2x + 3

  3. 2x + 4

  4. 2x + 5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
According to remainder theorem

$f(x)=\theta (x)(x^{2}-1)+\gamma (x)$

$\gamma (x)=ax+b$

So, $f(x)=\theta (x)(x^{2}-1)+(ax+b)$

$f(-1)=-a+b=3$      $\dots(1)$

$f(1)=a+b=7$         $\dots(2)$

Add $(1)$ and $(2)$

$-a+b+a+b=10\Rightarrow 2b=10\Rightarrow b=5$

substitute it in $(1)$ then

$a=5-b=5-3=2$ 

$\Rightarrow b=5; a=2$

So $\gamma (x)=2x+5$
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

Given $f(x)$ is a cubic polynomial in $x$. If $f(x)$ is divided by $(x + 3), (x + 4), (x + 5)$ and $(x + 6)$ then it leaves the remainders $0, 0, 4$ and $6$ respectively. Find the remainder when $f(x)$ is divided by $x + 7$.

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

It is given that $f(x)$ leaves the remainder $0$ if divided by $(x+3)$ and $(x+4)$, which implies that $(x+3)$ and $(x+4)$ are factors of $f(x)$.


Let the other factor be $ax+p$, then $f(x)$ is given by:

$f(x)=(x+3)(x+4)(ax+p)$

Now, it is also given that $f(x)$ leaves the remainders $4$ and $6$ if divided by $(x+5)$ and $(x+6)$, which means that $f(-5)=4$ and $f(-6)=6$.

If $f(-5)=4$, then we have:

$f(x)=(x+3)(x+4)(ax+p)\ \Rightarrow f(-5)=(-5+3)(-5+4)(a(-5)+p)\ \Rightarrow 4=(-2)(-1)(-5a+p)\ \Rightarrow 4=2(-5a+p)\ \Rightarrow -5a+p=2\quad ........(1)$

And if $f(-6)=6$, then we have:

$f(x)=(x+3)(x+4)(ax+p)\ \Rightarrow f(-6)=(-6+3)(-6+4)(a(-6)+p)\ \Rightarrow 6=(-3)(-2)(-6a+p)\ \Rightarrow 6=6(-6a+p)\ \Rightarrow -6a+p=1\quad ........(2)$

Subtract eqn 2 from eqn 1 as follows:

$[-5a-(-6a)]+(p-p)=2-1\ \Rightarrow (-5a+6a)+0=1\ \Rightarrow a=1$

Substitute the value of $a$ in eqn 1:

$(-5\times 1)+p=2\ \Rightarrow -5+p=2\ \Rightarrow p=2+5=7$

Therefore, 

$f(x)=(x+3)(x+4)[(1\times x+7)]\ \Rightarrow f(x)=(x+3)(x+4)(x+7)$

Thus, $f(-7)=0$

Hence, $f(x)$ leaves the remainder $0$ when divided by $x+7$.