### Polynomials

 Description: For classes 9 to 12 Number of Questions: 15 Created by: Bharat Dubey Tags:
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A trinomial is a polynomial with exactly how many terms?

1. 3

2. 4

3. 5

4. 6

Correct Option: A

What is called a polynomial with exactly 2 terms?

1. binomial

2. degree

3. linear

4. square

Correct Option: A

How many 0s can a polynomial of degree n have?

1. 1 zero

2. n zero

3. 2 zero

4. n-1 zero

Correct Option: B

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 $1$.

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 $g(x)$

If the polynomial $x^3-x^2+x-1$ is divided by $x-1$, then the quotient is :

1. $x^2-1$

2. $x^2+1$

3. $x^2-x+1$

4. $x^2+x+1$

Correct Option: B
Explanation:

Divide $x^3-x^2+x-1$ by   $x-1$

$x-1$ $\overline{)x^3-x^2+x-1(}$  $x^2+1$

$-(x^3-x^2)$
$\overline{\quad\quad\quad\quad+x-1}$
$-(x-1)$
$\overline{\quad\quad\quad0}$

Hence, $B$ is correct.

When the polynomial  ${x^4} + {x^2} + 1$   is divided by $(x + 1)({x^2} - x + 1)$ then the remainder is $ax + b$ , then  $a + b$ is equal to

1. $1$

2. $2$

3. $3$

4. $4$

Correct Option: A
Explanation:
$\dfrac{{x}^{4}+{x}^{2}+1}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

$=\dfrac{{x}^{4}+2{x}^{2}+1-{x}^{2}}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

$=\dfrac{{\left({x}^{2}+1\right)}^{2}-{x}^{2}}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

$=\dfrac{\left({x}^{2}-x+1\right)\left({x}^{2}+x+1\right)}{\left(x+1\right)\left({x}^{2}-x+1\right)}$

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

$=\dfrac{x\left(x+1\right)+1}{\left(x+1\right)}$

$=x+\dfrac{1}{x+1}$

Remainder$=1$ is of the form $ax+b$

$\Rightarrow\,a=0,\,b=1$

$\therefore\,a+b=0+1=1$

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder.
$p(x)=x^4-3x^2+4x+5$
$g(x)=x^2+1-x$

1. $q(x)=x^2+x-3$ and $r(x)=-8$

2. $q(x)=x^2-x+3$ and $r(x)=8$

3. $q(x)=x^2+x-3$ and $r(x)=8$

4. $q(x)=x^2-x-3$ and $r(x)=-8$

Correct Option: C
Explanation:

$x^2-x+1)\overline {x^4-3x^2+4x+5}$ ( $x^2+x-3$
$\underline {\underset {-}{}x^4\underset {-}{+}x^2 \underset{+}{-}x^3}$
$x^3-4x^2+4x+5$
$\underline {\underset {-}x^3\underset {+}{-}x^2\underset {-}{+}x}$
$-3x^2+3x+5$
$\underline {\underset {+}{-}3x^2\underset {-}{+}3x\underset {+}{-}3}$
$8$
Hence, Quotient=$x^2+x-3$
Remainder=8.

On dividing $f(x)$ by a polynomial $x-1-x^2$, the quotient $q(x)$ and remainder $r(x)$ are $(x-2)$ and $3$ respectively. Then $f(x)$ is

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

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

3. $f(x)=3x^2-3x+5$

4. $f(x)=-x^3+3x^2-3x+5$

Correct Option: D
Explanation:

$f(x)=q(x)g(x)+r(x)$

$\therefore f(x)= (x-2)(x-1-x^2)+3$

$\Rightarrow f(x)= x(x-1-x^2)-2(x-1-x^2)+3$

$=x^2-x-x^3-2x+2+2x^2+3$

$=-x^3+3x^2-3x+5$

On dividing $x^3-3x^2+x+2$ by a polynomial $g(x)$, the quotient and remainder were $(x-2)$ and $(-2x+4)$, respectively. Find $g(x)$.

1. $2x^2+2x-8$

2. $x^2+2x-7$

3. $x^2-x+1$

4. $2x^2-x+2$

Correct Option: C
Explanation:
By Remainder theorem,
$p(x)=g(x)q(x)+r(x)$

We have, $p(x)=x^3-3x^2+x+2,q(x)=x-2\space and \space r(x)=-2x+4$

$\therefore x^3-3x^2+x+2=g(x)(x-2)+(-2x+4)$

$\Rightarrow x^3-3x^2+x+2+2x-4=g(x)(x-2)$

$\Rightarrow g(x)=\dfrac{x^3-3x^2+3x-2}{(x-2)}=\dfrac{x^3-2x^2-x^2+2x+x-2}{(x-2)}$

$=\dfrac{[x^2(x-2)-x(x-2)+1(x-2)]}{(x-2)}$

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

$=x^2-x+1$

$\therefore g(x)=x^2-x+1$

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

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, $k=5,a=-5$

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 $x^3-4x^2-25x+70$.

The remainders of polynomial f(x) when divided by x-1, x-2 are 2,3 then the remainder of f(x) when divided by (x-1) (x-2) is

1. 2x-1

2. x-1

3. 2x+1

4. x+1

Correct Option: D
Explanation:
According to Remainder theorem

$f(x)=(x-1)(x-2) \theta (x)+\gamma (x)$

$\gamma (x)=ax+b$

$f(1)=a+b=2$     $\dots(1)$

$f(2)=2a+b=3$     $\dots(2)$

Subtract $(1)$ from $(2)$

$2a+b-a-b=3-2$

$\Rightarrow a=1$ substitute in $(1)$

$b=2-1=1$

$\therefore\ a=b=1$

So, $\gamma (x)=x+1$

If the remainders of the polynomial f(x) when divided by x+1 and x-1 are 3, 7 then the remainder of f(x) when divided by $(x^{2} -1 )$ is

1. x + 4

2. 2x + 3

3. 2x + 4

4. 2x + 5

Correct Option: D
Explanation:
According to remainder theorem

$f(x)=\theta (x)(x^{2}-1)+\gamma (x)$

$\gamma (x)=ax+b$

So, $f(x)=\theta (x)(x^{2}-1)+(ax+b)$

$f(-1)=-a+b=3$      $\dots(1)$

$f(1)=a+b=7$         $\dots(2)$

Add $(1)$ and $(2)$

$-a+b+a+b=10\Rightarrow 2b=10\Rightarrow b=5$

substitute it in $(1)$ then

$a=5-b=5-3=2$

$\Rightarrow b=5; a=2$

So $\gamma (x)=2x+5$
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