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

$p(x)=(x^2-10x-24)$ , when divided by $x+2$ and $x\neq -2$ gives the quotient $Q$. Find $Q$.

  1. $x -22$
  2. $x-12$
  3. $x+12$
  4. $x+22$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
 Quotient  $x - 12$
 $x+2$   $x^2-10x-24$  $x^2+2x$$- $   $-$
         $ -12x-24$        $ -12x-24$       $+$        $+$------------------------------                 $ 0$

Using division algorithm method we get the value of $Q = x - 12$.

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

When $(x^3-2x^2+px-q)$ is divided by $(x^2-2x-3)$, the remainder is $(x-6)$. The values of $p$ and $q$ respectively are ____. 

  1. $-2, 7$
  2. $2, -6$
  3. $-2, 6$
  4. $2, 6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$x^3-2x^2+px-q-(x-6)=0$ at $x=-1$ and $x=3$.         ($\because (x-6)$ is the remainder)
put $x=-1$,
$-1-2-p-q+1+6=0\ \Rightarrow p+q=4\dots eqn (1)$
Now, put $x=3$
$27-18+3p-q-3+6=0\ \Rightarrow 3p-q=-12\dots eqn (2)$
Add equation 1 and 2, we get
$4p=-8\Rightarrow p=-2$
 and $q=4-p=6$

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

When a polynomial $P(x)$ is divided by $x, (x - 2)$ and $(x - 3)$, remainders are $1$, $3$ and $2$ respectively. the same polynomial is divided by $x(x - 2)(x - 3)$, the remainder is $ax^2 + bx + c$, then the value of $c$ is

  1. $-3$
  2. $-2$
  3. $6$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to factor theorem

$p(x)=xq(x)+1$
$P(x)=(x-2)q'(x)+3$
$P(x)=(x-3)q''(x)+2$
So, At $x=0 , P(0)=1$
At $x=2 , P(2)=3$
At $x=3 , P(3)=2$
Now when the polynomial $P(x)$ is divided by $(x-2)(x-3)x$ the remainder must have the degree less than $3$ . that is the remainder will be of the form $ax^2+bx+c$
$\implies P(x)=x(x-2)(x-3)q''''(x)+ax^2+bx+c$
So, $P(0) =1=a(0)^2+b(0)+c\implies c=1$

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

The quotient and remainder when $x^{2002}$ $- 2001$ is divided by $x^{91}$ are 

  1. $x^{91 \times 22}, 2001$
  2. $x^{91}, 2001$
  3. $x^{91\times 21}, -2001$
  4. $x^9, -2001$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2002 = 91 \times 22$

$\therefore x^{2002} = x^{91 \times 22}$
$\therefore$ $x^{2002} - 2001$ $=$ $x^{91} \times (x^{91 \times 21}) - 2001$

When $x^{91} \times (x^{91 \times 21}) - 2001$ is divided by $x^{91}$, then
Quotient $= x^{91 \times 21}$ 
And 
Remainder $= -2001$

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

Which of the following given options is/ are correct?
If $p(x)=q(x)g(x)+r(x)$ (By Division Algorithm) where p(x), g(x) are any two polynomials with $g(x)\neq 0$, then

  1. $r(x)=0$ always

  2. degree of r(x)< degree of g(x) always
  3. either $r(x)=0$ or degree of r(x)< degree of g(x)
  4. $r(x)=g(x)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

(a) If p(x) is not divisible by g(x), then $r(x)\neq 0 \therefore$ (a) is not true
(b) If p(x) is divisible by g(x), the $r(x)=0$ for all x i.e., r(x) is zero polynomial whose degree is not defined.
$\therefore $(b) is not true
(c) is clearly true [$\because$ division algorithm rule]
Since degree of $r(x)<$ degree of g(x) or $r(x)=0$, but $g(x)\neq 0$.
(d) $\therefore r(x)=g(x)$ is not true.

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

A polynomial $f(x)$ with rational coefficient leaves reminder $15$, when divided by $(x-3)$ and remainder $2x+1$, when divided by $(x-1)^{2}$. If $p$ is coefficient of $x$ of its remainder which will come out if $f(x)$ is divided by $(x-3)(x-1)^{2}$ then find $p$.

  1. $-2$
  2. $-1$
  3. $1$
  4. $6$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the remainder be ax^2+bx+c. Using f(3)=15 and the condition for (x-1)^2, we set up a system of equations. Solving for the remainder when divided by (x-3)(x-1)^2 yields a linear term coefficient p = -2.

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

The remainder obtained when the polynomial $1+x+x^ {3}+x^ {9}+x^ {27}+x^ {81}+x^ {243}$ is divisible by $x-1$ is

  1. $3$
  2. $5$
  3. $7$
  4. $11$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$p{\left( x \right)} = 1 + x + {x}^{3} + {x}^{9} + {x}^{27} + {x}^{81} + {x}^{243}$

Let $q{\left( x \right)}$ be the quotient when $P{\left( x \right)}$ divided by $\left( x - 1 \right)$.
Therefore,
$P{\left( x \right)} = \left( x - 1 \right) \cdot q{\left( x \right)} + A$
$P{\left( 1 \right)} = \left( 1 - 1 \right) \cdot q {\left( x \right)} + A$
$7 = 0 + A$
$A = 7$
Hence the remainder is $7$.

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

If $A=2x^{3}+5x^{2}+4x+1$ and $B=2x^{2}+3x+1$, then find the quotient from the following four option, when A is divided by B.

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


$\dfrac{A}{B} = \dfrac{2x^{3}+5x^{2}+4x+1}{2x^{2}+3x+1}$

$=\dfrac{2x^{3}+(3x^{2}+2x^{2})+(x+3x)+1}{2x^{2}+2x+1}$

$=\dfrac{(2x^{3}+3x^{2}+x)+(2x^{2}+3x+1)}{2x^{2}+3x+1}$

$=\dfrac{(2x^{2}+3x+1)(x+1)}{2x^{2}+3x+1}=x+1$

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

Evaluate: $\displaystyle \frac{a^3\, +\, b^3\, +\, c^3\, -\, 3abc}{a^2\, +\, b^2\, +\, c^2\, -\, ab\, -\, bc\, -\, ca}$

  1. $0$
  2. $a + b + c$
  3. $1$

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

$\cfrac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca } \ =\cfrac { \left( a+b+c \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right)  }{ { (a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca) } \ =a+b+c$.

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

If x + 2 and x-1 are the factors of $x^3 + 10x^2+mx + n$, then the values of m and n are respectively

  1. 5 and -3

  2. 17 and -8

  3. 7 and-18

  4. 23 and -19
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $x + 2$ is a factor of $x^3 + 10x^2 + mx + n$
$x =-2$
$(-2)^3 + 10(-2)^2 + m(-2) + n =0$
$ -8 + 40 = 2m - n $
$2m -n = 32$                     .....(i)
Again, $x-1 $ is a factor of $x^3 + 10x^2 + mx + n$
$x =1$
$1+10+m+n=0$
$m + n =-11$                    .....(ii)
Adding (i) and (ii). we get,
$3m = 21$
$m=7$
By putting m in (i). we get,
$2(7) - n = 32 $
$ - n = 18 $
   $n = -18$
Option C is correct.