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

The expression that should be subtracted from $\displaystyle 4x^{4}-2x^{3}-6x^{2}+x-5$ so that is may be exactly divisible by $\displaystyle 2x^{2}+x-2$ is

  1. $\displaystyle 3x+5$
  2. $\displaystyle -3x-5$
  3. $\displaystyle -3x+5$
  4. $\displaystyle 3x-5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
 $2x^2-2x$
$2x^2+x-2$  $4x^4-2x^3-6x^2+x-5$ $4x^4+2x^3-4x^2$
        $-4x^3-2x^2+x$      $-4x^3-2x^2+4x$
                             $-3x-5$

Thus, $-3x-5$ must be subtracted to make it exactly divisble

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

For a polynomial, dividend is $\displaystyle x^{4}+4x-2x^{2}+x^{3}-10$, quotient is $\displaystyle x^{2}+3x-3x^{2}+4x+12$ and remainder is $14$, then divisor is equal to

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

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

$p(x)=x^4+4x-2x^2+x^3-10$ and $q(x)=x^2+3x-3x^2+4x+12$ and $r(x)=14$

$p(x)-14$ is not divisible by $(x^2+2),(x^2-2),(x+2)$  
hence correct option is (D)

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

If $\displaystyle \left ( x^{2}+4x-21 \right )$ is divided by  $x + 7$  then the quotient is

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

Consider the polynomial $f(x)=x^2+4x-21$ and factorise it as follows:


$f(x)=x^2+4x-21=(x^2+4x+4)-21-4=(x+2)^2-25=(x+2)^2-(5)^2=(x+2+5)(x+2-5)$
$=(x+7)(x-3)$

Therefore, $f(x)=(x+1)(x-1)(x-3)$

Let $g(x)=x+7$

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

$q(x)=\frac { (x+7)(x-3) }{ (x+7) } =x-3$

Hence, the quotient is $x-3$.

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

When a number is divided by $13$, the remainder is $11$. When the same number is divided by $17$, the remainder is $9$. What is the number ?

  1. $853$
  2. $278$
  3. $349$
  4. $670$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x = 13p + 11$ and $x = 17q + 9$
$\therefore$ $13p + 11 = 17q + 9$
$\therefore$ $17q - 13p = 2$
$\therefore$ q $=\dfrac{2 + 13p}{17}$
The least value of p for which q $=\dfrac{2 + 13p}{17}$ is a whole number is $p = 26$
x $= (13 \times 26 + 11)$
$= (338 + 11)$
$= 349$

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

On dividing a number by $56$, we get $29$ as remainder. On dividing the same number by $8$, what will be the remainder ?

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

Applying remainder theorem i.e. $A=bq+r $
where, $b$ = divisior 
$r$ = remainder 
$\therefore A = 56q + 29$
if q = 1 
The no. is $A = 85$
On dividing by $8$, Remainder $(r) =5$

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 $x+7$.

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

Find the remainder when  $-2x^3-2x^2+27x-30$ is divided by $2-x$.

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Quotient----->  $2x^2+6x-15$
 -x + 2   $-2x^2-2x^2+27x-30$  $-2x^3+4x^2$
    $-6x^2+27x-30$ $-6x^2+12x$-----------------------------------$15x-30$$15x-30$-------------------0