Tag: remainder and factor theorems

Questions Related to remainder and factor theorems

Polynomials $p(x), g(x), q(x)$ and $r(x)$, which satisfy the division algorithm and "deg $q(x) = $ deg $ r(x)$", are

  1. $p(x)=2x^2+x; g(x)=2x^2-4$;
    $q(x)=2x-7; r(x)=-x+2$

  2. $p(x)=x^2+x-3; g(x)=x^2+x-1$;
    $q(x)=7; r(x)=-5$

  3. $p(x)=x^2+x; g(x)=x^2-4$;
    $q(x)=2x-1; r(x)=-x-2$

  4. $p(x)=2x^2+2x+8; g(x)=x^2+x+9$;
    $q(x)=2; r(x)=-10$


Correct Option: D
Explanation:

according to division algorithm $p(x)=q(x)g(x)+r(x)$

degree of $q(x)$ is equal to $r(x)$ in all options.
only (D) option satisfies $p(x)=q(x)g(x)+r(x)$
$g(x)q(x)=2(x^2+x+9)=2x^2+2x+18=p(x)+10=p(x)-r(x)$ hence $p(x)=q(x)g(x)+r(x)$ in (D) satisfies division algorithm

On dividing $f(x)=2x^5+3x^4+4x^3+4x^2+3x+2$ by a polynomial $g(x)$, where $g(x)=x^3+x^2+x+1$, the quotient obtained as $2x^2+x+1$. Find the remainder $r(x)$.

  1. $r(x)=7x^3+x^2-1$

  2. $r(x)=3x^2+2x+1$

  3. $r(x)=x-2$

  4. $r(x)=x+1$


Correct Option: D
Explanation:

By remainder theorem,
$f(x)=q(x)g(x)+r(x)$
$\therefore 2x^5+3x^4+4x^3+4x^2+3x+2=(2x^2+x+1)(x^3+x^2+x+1)+r(x)$
$=2x^2(x^3+x^2+x+1)+x(x^3+x^2+x+1)+1(x^3+x^2+x+1)+r(x)$
$=2x^5+2x^4+2x^3+2x^2+x^4+x^3+x^2+x+x^3+x^2+x+1+r(x)$
$=2x^5+3x^4+4x^3+4x^2+2x+1+r(x)$
$r(x)=x+1$

Polynomials $p(x), g(x), q(x)$ and $r(x)$, which satisfy the division algorithm and "deg $p(x) = $ deg $q(x)$" are

  1. $p(x)=2x^2+2x+8, g(x)=4x+1$;
    $q(x)=x^2; r(x)=1$

  2. $p(x)=2x^2+2x+8, g(x)=5$;
    $q(x)=4; r(x)=4x-1$

  3. $p(x)=2x^2+2x+8, g(x)=2$;
    $q(x)=x^2+x+4; r(x)=0$

  4. $p(x)=x^2+x+3, g(x)=2x+3$;
    $q(x)=2x^2+x; r(x)=3x-2$


Correct Option: C
Explanation:

degree of $p(x)$ and $q(x)$ are equal in (A),(C),(D)

according to division algorithm, $p(x)=q(x)g(x)+r(x)$
in option (C), $g(x)q(x)=2(x^2+x+4)=2x^2+2x+8+0=g(x)q(x)+r(x)=p(x)$
hence option (C) is correct answer.

What should be added to $8x^4+14x^3-2x^2+7x-8$ so that the resulting polynomial is exactly divisible by $4x^2+3x-2$?

  1. $10-14x$

  2. $4x-10$

  3. $3x-5$

  4. $5-3x$


Correct Option: A
Explanation:

$4x^2+3x-2)\overline {8x^4+14x^3-2x^2+7x-8}$ ( $2x^2+2x-1$
                          $\underline {\underset {-}{8}x^4\underset {-}{+}6x^3\underset {+}{-}4x^2}$
                                     $8x^3+2x^2+7x-8$
                                     $\underline {\underset {-}{8}x^3\underset {-}{+}6x^2\underset {+}{-}4x}$
                                             $-4x^2+11x-8$
                                             $\underline {\underset {+}{-}4x^2\underset {+}{-}3x\underset {-}{+}2}$
                                                          $14x-10$
We have to add $10-14x$ so that $8x^4+14x^3-2x^2+7x-8$ is completely divisible by $4x^2+3x-2$.

Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm. $x^3-3x+1, x^5-4x^3+x^2+3x+1$

  1. Yes

  2. No

  3. Ambiguous

  4. Data insufficient


Correct Option: B
Explanation:

$x^3-3x+1)\overline {x^5-4x^3+x^2+3x+1}$($x^2-1$
                         $\underset {-}{x^5}\underset {+}{-}3x^3\underset {-}{+}x^2$
                         $\overline {-x^3+3x+1}$
                         $\underline {\underset {+}{-}x^3\underset {-}{+}3x\underset {+}{-}1}$
                                              $2$
Since remainder is non-zero.
Therfore,$x^3-3x+1$ is not a factor of $x^5-4x^3+x^2+3x+1$

If the polynomial $f(x)=x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the remainder comes out to be $(x+a)$, then values of $k$ and $a$ are

  1. $k=-2$ & $a=4$

  2. $k=5$ & $a=-5$

  3. $k=-3$ & $a=-7$

  4. None of these


Correct Option: B
Explanation:

$f(x)=$ is divided by another polynomial
$x^2-2x+k)\overline {x^4-6x^3+16x^2-25x+10}(x^2-4x+(8-k)$
                       $\underline {\underset {-}{x^4}\underset {+}{-2x^3}\underset{-}{+}kx^2}$
                       $-4x^3+(16-k)x^2-25x+10$
                       $\underline {\underset {-}{-4x^3}\underset {-}{+8x^2}                      \underset{+}{-}4kx}$
                       $(8-k)x^2+(4k-25)x+10$
                       $\underline {\underset {-}(8-k)x^2+\underset {-}(2k-16)x\underset{-}{+}(8k-k^2)}$
                       $(2k-9)x+(k^2-8k+10)$
But remainder is given $x+a$
$\therefore x+a=(2k-9)x+(k^2-8k+10)$
On equating coefficient, we get
$2k-9=1\Rightarrow k=5$
and $a=k^2-8k+10\Rightarrow a=25-40+10=-5$
Hence, $k=5,a=-5$

Find the value of $b$ for which the polynomial $2x^3+9x^2-x-b$ is exactly divisible by $2x+3$?

  1. $15$

  2. $-15$

  3. $10$

  4. $-10$


Correct Option: A
Explanation:

Since $2x+3$ is a factor of the polynomial $p\left(x\right)=2x^3+9x^2-x-b$
Therefore, by Factor theorem $p\left(-\dfrac32\right)=0$
$\Rightarrow 2\left(-\dfrac32\right)^3+9\left(-\dfrac32\right)^2-\left(-\dfrac32\right)-b=0$

$\Rightarrow -\dfrac{27}4+\dfrac{81}4+\dfrac32-b=0$

$\Rightarrow \dfrac{-27+81+6}4-b=0\Rightarrow b=\dfrac{60}4=15$

$\therefore \space b=15$

What must be subtracted from or added to $8x^4+14x^3-2x^2+8x-12$ so that it may be exactly divisible by $4x^2+3x-2$?

  1. $15x-14$

  2. $3x-14$

  3. $-15x+14$

  4. $-3x+14$


Correct Option: C
Explanation:

$4x^2+3x-2)\overline {8x^4+14x^3-2x^2+8x-12}$ ( $2x^2+2x-1$
                            $\underline {\underset {-}{8}x^4\underset {-}{+}6x^3\underset {+}{-}4x^2}$
                            $8x^3+2x^2+8x-12$
                            $\underline {\underset {-}{8}x^3\underset {-}{+}6x^2\underset {+}{-}4x}$
                            $-4x^2+12x-12$
                            $\underline {\underset {+}{-}4x^2\underset {+}{-}3x\underset {-}{+}2}$
                                        $15x-14$

$\therefore$ The expression that must be subtracted is $15x-14$
and the expression that must be added is $-(15x-14)=-15x+14$

$\displaystyle \left( { 3x }^{ 2 }-x \right) \div \left( -x \right) $ is equal to

  1. $\displaystyle 3x+1$

  2. $\displaystyle -3x-1$

  3. $\displaystyle -3x+1$

  4. $\displaystyle 3x-1$


Correct Option: C
Explanation:

$\displaystyle \left( { 3x }^{ 2 }-x \right) \div \left( -x \right)$

By separating denominators, we get
$  =\dfrac { 3{ x }^{ 2 } }{ -x } +\dfrac { \left( -x \right)  }{ \left( -x \right)  } =-3x+1$
Hence, final result after given operation is $-3x+1$.

A polynomial when divided by $\displaystyle \left ( x-6 \right )$ gives a quotient $\displaystyle x^{2}+2x-13$ and leaves a remainder $-8$. Then polynomial is

  1. $\displaystyle x^{3}+4x^{2}+25x-78$

  2. $\displaystyle x^{3}-4x^{2}-25x+70$

  3. $\displaystyle x^{3}-4x^{2}-25x-70$

  4. $\displaystyle x^{3}+4x^{2}-25x+78$


Correct Option: B
Explanation:
Let $P$ be the polynomial. If $P$ is divided by $(x-6)$ then it leaves a remainder $-8$ and gives a quotient $x^2+2x-13$. Therefore, 

$\cfrac { P }{ x-6 } ={ x }^{ 2 }+2x-13-\cfrac { 8 }{ x-6 } \\ \Rightarrow P=(x-6)({ x }^{ 2 }+2x-13)-\frac { 8(x-6) }{ x-6 } \\ \Rightarrow P={ x }^{ 3 }+2{ x }^{ 2 }-13x-6{ x }^{ 2 }-12x+78-8\\ \Rightarrow P={ x }^{ 3 }-4{ x }^{ 2 }-25x+70$

Hence, the polynomial is $x^3-4x^2-25x+70$.