Tag: division algorithm for polynomials

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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

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

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

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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$=f\left(-\pi\right)={\left(-\pi\right)}^{3}+3{\left(-\pi\right)}^{2}+3\left(-\pi\right)+1=-{\pi}^{3}+3{\pi}^{2}-3\pi+1$
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 $ax+b$

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

$\therefore\,a+b=0+1=1$
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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

  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 $r(x)=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $r(x)=0$.

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

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

  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 $r(x)=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $r(x)=0$.


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

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$

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient
Reveal answer Fill a bubble to check yourself
A Correct answer
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).                     
                                             

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

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

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

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

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

$=x^2-x+1$

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