Tag: linear and synthetic method of division

Quizzes (0) Questions (64)

Questions Related to linear and synthetic method of division

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

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 $(x-1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For f(x) to be divisible by x^2+x+1, it must vanish at the roots of x^2+x+1=0, which are omega and omega^2. Substituting x=omega into P(x^3)+Q(x^3) gives P(1)+Q(1)=0, implying P(1)=-Q(1). This condition is satisfied if both P(x) and Q(x) contain the factor (x-1).

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

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. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $1$.
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(2x+3)$

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

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. $3x^3\, +\, 4x^2\, -\, x\, -\,3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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. $\cfrac{a+b+cd}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $-1,-4$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 
$a=-1, b=-4$

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\boxed{3}$

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Then, deg $p(x) \ge$ deg $g(x)$