Tag: remainder and factor theorems

Questions Related to remainder and factor theorems

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

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

  1. $\displaystyle 3x+5$

  2. $\displaystyle -3x-5$

  3. $\displaystyle -3x+5$

  4. $\displaystyle 3x-5$

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Correct Option: B
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

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

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

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

  2. $\displaystyle x^{2}-2$

  3. $\displaystyle x+2$

  4. None of these

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Correct Option: D
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)

If $\displaystyle \left ( x^{2}+4x-21 \right )$ is divided by  $x + 7$  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. $\displaystyle x+3$

  2. $\displaystyle x-3$

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

  4. $\displaystyle x-4$

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Correct Option: B
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 $x-3$.

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 

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

  1. $7$

  2. $8$

  3. $9$

  4. $10$

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Correct Option: D
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.

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 ?

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

  1. $853$

  2. $278$

  3. $349$

  4. $670$

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Correct Option: C
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 $= (13 \times 26 + 11)$
$= (338 + 11)$
$= 349$

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

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

  1. $4$

  2. $5$

  3. $6$

  4. $7$

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Correct Option: B
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 $(r) =5$

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

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

  1. 2x-1

  2. x-1

  3. 2x+1

  4. x+1

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Correct Option: D
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, $\gamma (x)=x+1$

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

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

  1. x + 4

  2. 2x + 3

  3. 2x + 4

  4. 2x + 5

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Correct Option: D
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 $\gamma (x)=2x+5$

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 $x + 7$.

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

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Correct Option: A
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 $x+7$.

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

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: A
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