Tag: remainder and factor theorems

Questions Related to remainder and factor theorems

If $P(x)$ and $Q(x)$ are two polynomial such that $f(x)=P(x^3)+Q(x^3)$ is divisible by $x^2+x+1$, then?

  1. $P(x)$ is divisible by $(z-1)$ by $Q(x)$ is not divisible by $(x-1)$

  2. $Q(x)$ is divisible by $(x-1)$ but $P(x)$ is not divisible by $(x-1)$

  3. Both $P(x)$ and $Q(x)$ are divisible by $(x-1)$

  4. $f(x)$ is divisible by $(x-1)$


Correct Option: C

What is the degree of the remainder atmost, when a fourth degree polynomial is divided by a quadratic polynomial?

  1. $2$

  2. $0$

  3. $4$

  4. $1$


Correct Option: D
Explanation:
Here $f(x)$ represent dividend and $g(x)$ represent division

$g(x)=$ quadratic polynomial $=ax^2+bx+c$

$\therefore deg(g(x))=2$, $deg(f(x))=4$

quotient $q(x)$ is of degree $=2$ $(=4-2)$

Remainder $R(x)=$ degree $1$ or less than $1$.

Can $(x - 1)$ be the remainder on division of a polynomial $p(x)$ by $2x + 3$?

  1. Yes

  2. No

  3. Cannot be justified

  4. None of above


Correct Option: B
Explanation:

Division algorithm stated that a polynomial $f(x)$ can written as
   $f(x) = g(x)q + r$     where $q$ and $r$ are unique integer and $0 <= r < g(x)$.
Here,
$g(x)=2x+3$ and $r(x)=x-1$
The power of the remainder is always less than the power of the divisor
Here, the degree of remainder is $1$ and the degree of divisor is $1$, which is not possible. Thus, $(x-1)$ cannot be the remainder of $p(x)$ when divided by $(2x+3)$

If quotient = $3x^2\, -\, 2x\, +\, 1$, remainder = $2x - 5$ and divisor  = $x + 2$, then the dividend is:

  1. $3x^3\, -\,4x^2\, +\, x\, -\, 3$

  2. $3x^3\, -\, 4x^2\, -\, x\, +\, 3$

  3. $3x^3\, +\, 4x^2\, -\, x\, +\, 3$

  4. $3x^3\, +\, 4x^2\, -\, x\, -\,3$


Correct Option: D
Explanation:

Dividend = (divisor $\times$ quotient) + remainder
= $(3x^2\, -\, 2x\, +\, 1)\, \times\,  (x\, +\, 2)\,+\, (2x\, -\, 5)$
= $3x^3\, +\, 4x^2\, -\, x\,-\, 3$

The remainder if $a{x}^{3}+b{x}^{2}+cx+d$ is divided by $ax+b$

  1. $ad-bc$

  2. $\cfrac{1}{a}(ad-bc)$

  3. $\cfrac{a-bc}{d}$

  4. $\cfrac{a+b+cd}{2}$


Correct Option: B
Explanation:

Here, $p(x) = a{x}^{3}+b{x}^{2}+cx+d$ and factor of $ax+b$ is
$ax+b = 0$
$x = -\frac ba$
$ p(-\frac ba) = a(-\frac ba)^{3}+b(-\frac ba)^{2}+c(-\frac ba)+d$
$p(-\frac ba) = -\frac {b^3}{a^2}+ \frac {b^3}{a^2} -\frac {bc}{a}+d $
$p(-\frac ba) =  -\frac {bc}{a}+d $
$p(-\frac ba) = \frac 1a (ad - bc) $
when $a{x}^{3}+b{x}^{2}+cx+d$ is divided by $ax+b$ then remainder is $p(-\frac ba) = \frac 1a (ad - bc) $
Option B is correct.

If $ \displaystyle 2x^{3}+4x^{2}+2ax+b $  is exactly divisible by $ \displaystyle x^{2}-1 $  Then the value of $a$ and $b$ respectively will be 

  1. $1,2$

  2. $-1,4$

  3. $1,-2$

  4. $-1,-4$


Correct Option: D
Explanation:

Since $\displaystyle f\left ( x \right )=2x^{3}+4x^{2}+2ax+b$ is exactly divisible
by $x\displaystyle ^{2}-1=\left ( x-1 \right )\left ( x+1 \right )$
$\displaystyle \therefore f\left ( 1 \right )=0$ and $\displaystyle f\left ( -1 \right )=0$
These give
    $2+4+2a+b=0$
or $2a+b+6=0$          .....(i)
and $-2+4-2a+b=0$
or $2a-b-2=0$        ....(ii)
Solving equations (i) and (ii) we get 
$a=-1, b=-4$

The product of $x^2y$ and $\cfrac{x}{y}$ is equal to the quotient obtained when $x^2$ is divided by ____.

  1. $0$

  2. $1$

  3. $x$

  4. $\cfrac{1}{x}$


Correct Option: D
Explanation:

$x^2y\times\dfrac{x}{y}=x^3\Rightarrow $when $x^2$ is divided by$ \dfrac{1}{x}\ $gives $  x^3$

There is a remainder of 3 when a number is divided by 6. What will be the remainder if the square of the same number is divided by 6?

  1. 1

  2. 0

  3. 3

  4. 2


Correct Option: C
Explanation:

Let the no. be $'x'$ $\Rightarrow x=6q+3$

$x^{2}=(6q+3)^{2}=36q^{2}+36q+9$
Divide $x^{2}$ by $6$ 
we get $\boxed{x^{2}=6(6q^2+6q+1)+3}$
Hence the remainder is $\boxed{3}$

If on dividing a non-zero polynomial $p(x)$ by a polynomial $g (x)$, the remainder is zero, what is the relation between the degrees of $p(x)$ and $g (x)$?

  1. degree of $g (x) \ge$ degree of $p(x)$

  2. degree of $g(x) \le$ degree of $p(x)$

  3. degree of $g (x) =$ degree of $p(x)$

  4. Can't say


Correct Option: B
Explanation:

deg $p(x)=$ deg $g(x)+r(c)$

Then, deg $p(x) \ge$ deg $g(x)$