Tag: maths

Questions Related to maths

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?

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

  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 $(x-1)$

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Correct Option: C

The remainder when $3x^{4}+5x^{2}+9$ is divided by $x^{2}-2$ is

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

  1. $28$

  2. $13$

  3. $31$

  4. $17$

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Correct Option: C

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

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

  1. $2$

  2. $0$

  3. $4$

  4. $1$

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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 $2x + 3$?

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

  1. Yes

  2. No

  3. Cannot be justified

  4. None of above

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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:

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

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

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

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

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

Show answer
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 $ax+b$

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

  1. $ad-bc$

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

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

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

Show answer
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 

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

  1. $1,2$

  2. $-1,4$

  3. $1,-2$

  4. $-1,-4$

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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 ____.

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. $x$

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

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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?

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

  1. 1

  2. 0

  3. 3

  4. 2

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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 $g (x)$?

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

  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

Show answer
Correct Option: B
Explanation:

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

Then, deg $p(x) \ge$ deg $g(x)$