Tag: maths

Questions Related to maths

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

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

  1. $x^2-1$

  2. $x^2+1$

  3. $x^2-x+1$

  4. $x^2+x+1$

Show answer
Correct Option: B
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.

Find the reminder when ${x^3} + 3{x^2} + 3x + 1$ is divided by $x + \pi $

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

  1. $\pi$

  2. $-{\pi}^{2}+3{\pi}^{3}-3\pi+1$

  3. $-{\pi}^{3}+3{\pi}^{2}-3\pi+1$

  4. None of these

Show answer
Correct Option: C
Explanation:
The remainder theorem states that when a polynomial, $f\left(x\right)$, is divided by a linear polynomial , $x - a$, the remainder of that division will be equivalent to $f\left(a\right)$.

Given:$f\left(x\right)={x}^{3}+3{x}^{2}+3x+1$

$f\left(x\right)$ is divided by a linear polynomial , $x+\pi$, the remainder of that division will be equivalent to $f\left(-\pi\right)$.

Remainder$=f\left(-\pi\right)={\left(-\pi\right)}^{3}+3{\left(-\pi\right)}^{2}+3\left(-\pi\right)+1=-{\pi}^{3}+3{\pi}^{2}-3\pi+1$

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 

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: A
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 $ax+b$

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

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

Find the quotient $q(x)$ and remainder $r(x)$ of the following when $f(x)$ is divided by $g(x)$.
$p(x)=x^3-3x^2-x+3$;
$g(x)=x^2-4x+3$

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

  1. $q(x)=x-1$ and $r(x)=0$

  2. $q(x)=x+1$ and $r(x)=0$

  3. $q(x)=2x+1$ and $r(x)=0$

  4. $q(x)=2x-1$ and $r(x)=0$

Show answer
Correct Option: B
Explanation:

Consider the polynomial $f(x)=x^3-3x^2-x+3$ and factorise it as follows:


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

Therefore, $f(x)=(x+1)(x-1)(x-3)$

Now consider the polynomial $g(x)=x^2-4x+3$ and factorise it as follows:

$g(x)=x^2-4x+3=x^2-3x-x+3=x(x-3)-1(x-3)=(x-1)(x-3)$

Therefore, $g(x)=(x-1)(x-3)$


Now divide $f(x)$ by $g(x)$ to get $q(x)$:

$q(x)=\frac { (x-1)(x+1)(x-3) }{ (x-1)(x-3) } =x+1$

Since $f(x)$ is divisible by $g(x)$, therefore, the remainder $r(x)=0$.

Hence, the quotient $q(x)=x+1$ and remainder $r(x)=0$.

Find the quotient $q(x)$ and remainder $r(x)$ of the following when $f(x)$ is divided by $g(x)$.
$p(x)=x^6+x^4-x^2-1$;
$g(x)=x^3-x^2+x-1$

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

  1. $q(x)=2x^3+x^2+x-1$ and $r(x)=0$

  2. $q(x)=x^3+x^2+x-1$ and $r(x)=0$

  3. $q(x)=2x^3+x^3+x+1$ and $r(x)=0$

  4. $q(x)=x^3+x^2+x+1$ and $r(x)=0$

Show answer
Correct Option: D
Explanation:

Consider the polynomial $f(x)=x^6+x^4-x^2-1$ and factorise it as follows:


$f(x)=x^6+x^4-x^2-1=x^4(x^2+1)-1(x^2+1)=(x^4-1)(x^2+1)=(x^2-1)(x^2+1)(x^2+1)$
$=(x-1)(x+1)(x^2+1)^2$

Therefore, $f(x)=(x-1)(x+1)(x^2+1)^2$

Now consider the polynomial $g(x)=x^3-x^2+x-1$ and factorise it as follows:

$g(x)=x^3-x^2+x-1=x^2(x-1)+1(x-1)=(x^2+1)(x-1)$

Therefore, $g(x)=(x-1)(x^2+1)$


Now divide $f(x)$ by $g(x)$ to get $q(x)$:

$q(x)=\frac { (x-1)(x+1)(x^{ 2 }+1)^{ 2 } }{ (x-1)(x^{ 2 }+1) } =(x+1)(x^{ 2 }+1)=x^{ 3 }+x^{ 2 }+x+1$

Since $f(x)$ is divisible by $g(x)$, therefore, the remainder $r(x)=0$.

Hence, the quotient $q(x)=x^{ 3 }+x^{ 2 }+x+1$ and remainder $r(x)=0$.


Check whether $g(y)$ is a factor of $f(y)$ by applying the division algorithm.
$f(y)=3y^4+5y^3-7y^2+2y+2$
$ g(y)=y^2+3y+1$

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

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient

Show answer
Correct Option: A
Explanation:

Given f(y)=$3y^{4}+5y^{3}-7y^{2}+2y+2$ and (g)=$ y^{2}-4y+2$

Then $ y^{2}+3y+1\div 3y^{4}+5y^{3}-7y^{2}+2y+2\setminus 3y^{2}-y+7$
                                   $3y^{4}+9y^{3}+3y^{2}$
                                      -          -           -
                                  ---------------------------------------
                                       $-4y^{3}-10y^{2}+2y+2$
                                        $-4y^{3}-12y^{2}-4y+2$
                                            +          +           +
                                     -----------------------------------------
                                            $2y^{2}+6y+2$
                                             $2y^{2}+6y+2$
                                                -          -       -
                                    --------------------------------------------
                                                            0
So g(Y) is a factor of f(y).                     
                                             

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

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

  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 $r(x)=-8$

Show answer
Correct Option: C
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.

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

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

  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

Show answer
Correct Option: C
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).

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

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

  1. $f(x)=-3x^2-x+7$

  2. $f(x)=-x^3+x^2-x+7$

  3. $f(x)=3x^2-3x+5$

  4. $f(x)=-x^3+3x^2-3x+5$

Show answer
Correct Option: D
Explanation:

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

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

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

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

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

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 $g(x)$.

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

  1. $2x^2+2x-8$

  2. $x^2+2x-7$

  3. $x^2-x+1$

  4. $2x^2-x+2$

Show answer
Correct Option: C
Explanation:
By Remainder theorem,
$p(x)=g(x)q(x)+r(x)$

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

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

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

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

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

$=x^2-x+1$

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