Maths M.C.Q

$P(x)=5 x-4 x^2+3$
$\Rightarrow p(-1)=5(-1)-4(-1)^2+3$
$=-5-4+3$
$=-6$

A polynomial of degree $1$ is called a linear polynomial.
Options $(a)$, and $(c)$ have degree $2$,
so ther are quadratic polynomials.
option $(d)$ has a negative power, so it is not a polynomial.
The degree of $x + 1$ is $1$, so it is a linear polynomial.

$p(x)=x^3-a x^2+x$
$x-a=0 \Rightarrow x=a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x-a)$, the remainder is $p(a)$.
Now, $p(a)=a^3-a x^2+a$
$=a^3-a^3+a$
$=a$

$p(x)=x^4+2 x^3-3 x^2+x-1$
$x-2=0 \Rightarrow x=2$
By the remainder theorem, we know that when $\mathrm{p}(\mathrm{x})$ is divided by
$(x-2)$, the remainder is $p(2)$.
Now, $p(2)=x^4+2 x^3-3 x^2+x-1$
$=(2)^4+2(2)^3-3(2)^2+2-1$
$=16+16-12+2-1$
$=21$

$p(x)=2 x^2+7 x-4$
$\text { Now, } p(x)=0$
$\Rightarrow 2 x^2+7 x-4=0$
$\Rightarrow 2 x^2+8 x-x-4=0$
$\Rightarrow 2 x(x+4)-1(x+4)=0$
$\Rightarrow(x+4)(2 x-1)=0$
$\Rightarrow x+4=0 \text { and } 2 x-1=0$
$\Rightarrow x=-4 \text { and } x=\frac{1}{2}$

Let $f(x)=x^3+10 x^2+m x+n$
Now, $x+2=0 \Rightarrow x=-2$
and $\mathrm{x}-1=0 \Rightarrow \mathrm{x}=1$
By factor theorem,
$f(-2)=0$
$\Rightarrow(-2)^3+10(-2)^2+m(-2)+n$
$\Rightarrow-8+40-2 m+n=0$
$\Rightarrow 2 m-n=32 \ldots \text { (i) }$
By factor theorem,
$\mathrm{f}(1)=0$
$\Rightarrow(1)^3+10(1)^2+\mathrm{m}(1)+\mathrm{n}=0$
$\Rightarrow \mathrm{~m}+\mathrm{n}=-11 \ldots \text { (ii) }$
Adding $(i)$ and $(ii)$, we get
$3 m=21$
$\Rightarrow m=7$
Substituting in $(ii)$, we get
$n=-18$

A polynomial consisting of one term, namely zero only, is called a zero polynomial.
So, a zero polynomial can be defined as $p(x) = 0$.
This can also be written as $p(x)=0=0 x=0 x^2=0 x^3$ and so on.
So, it is not possible to determine the degree.
Hence, the degree of a zero polynomial is not defined.

$p(x)=x^3-3 x^2+4 x+32$
$x+2=0 \Rightarrow x=-2$
By the renainder theorem, we know that when $p(x)$ is divided by $(x+2)$, the remainder is $p(-2)$.
$\text { Now, } p(-2)=x^3-3 x^2+4 x+32$
$=(-2)^3-3(-2)^2+4(-2)+32$
$=-8-12-8+32$
$=4$

A polynomial with two non-zero terms is called a binomial.
$x^2+4$ is the polynomial that has two non-zero terms.
Hence is a binomial.

Let $\mathrm{p}(\mathrm{x})=2 \mathrm{x}^2+\mathrm{kx}$
Since $(x+1)$ is a factor of $p(x)$,
$=P(-1)=0$
$\Rightarrow 2(-1)^2+\mathrm{k}(-1)=0$
$\Rightarrow 2-\mathrm{k}=0$
$\Rightarrow \mathrm{k}=2$

Let $\mathrm{p}(\mathrm{x})$ be a polynomial. If $\mathrm{p}(\alpha)=0$, then we say that $\alpha$ is a zero of a polynomial.
$p(x)=2 x^2+5 x-3$
Now, $p(x)=0$
$\Rightarrow 2 x^2+5 x-3=0$
$\Rightarrow 2 x^2+6 x-x-3=0$
$\Rightarrow 2 x(x+3)-1(x+3)=0$
$\Rightarrow(2 x-1)(x+3)=0$
$\Rightarrow(2 x-1)=0 \text { or }(x+3)=0$
$\Rightarrow x=\frac{1}{2} \text { or } x=-3$
$\therefore \frac{1}{2}$ and $-3$ are the zeroes of the polynomial $p(x)$.

$p(x)=x^{100}+2 x^{99}+k$
$x+1=0 \Rightarrow x=-1$
By the factor theorem, we know that when $p(x)$ is divided by $(x+1)$, the remainder is $p(-1)$.
Now, $p(-1)$ $=(-1)^{100}+2(-1)^{99}+k$
$\Rightarrow 0=1-2+k \ldots($ Given that $\mathrm{p}(\mathrm{x})$ is divisible by $\mathrm{x}+1$.
$\Rightarrow \mathrm{k}=1$

$p(x)=3 x^2-1$
$\text { Now, } p(x)=0$
$\Rightarrow 3 x^2-1=0$
$\Rightarrow 3 x^2=1$
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}\ \text{and}\ -\frac{1}{\sqrt3}$

$p(x)=x^3+a x^2+2 x+a$
$x+a=0 \Rightarrow x=-a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x+a)$, the remainder is $p(-a)$.
Now, $p(-a)=x^3+a x^2+2 x+a$
$=(-a)^3+a(-a)^2+2(-a)+a$
$=-a^3+a^3-2 a+a$
$=-a$

$p(x)=x^3-20 x+5 k$
Now, $x+5=0 \Rightarrow x=(-5)$
By factor theorem,
$p(-5)=0$
$\Rightarrow(-5)^3-20(-5)+5 k=0$
$\Rightarrow-125+100+5 k=0$
$\Rightarrow-25+5 k=0$
$\Rightarrow 5 k=25$
$\Rightarrow k=5$

Let $\mathrm{p}(\mathrm{x})$ be a polynomial. If $\mathrm{p}(\alpha)=0$, then we say that $\alpha$ is a zero of a polynomial.
$p(x)=x^2+x-6$
Now, $\mathrm{p}(\mathrm{x})=0$
$\Rightarrow x^2+x-6$
$\Rightarrow x^2+3 x-2 x-6=0$
$\Rightarrow x(x+3)-2(x+3)=0$
$\Rightarrow(x-2)(x+3)=0$
$\Rightarrow(x-2)=0 \text { or }(x+3)=0$
$\Rightarrow x=2 \text { or } x=-3$
$\therefore 2$ and $-3$ are the zeroes of the polynomial $p(x)$.

Given, $x^3+2 x^2-x-2$
For $f(-1)$,
$-1+2(-1)^2-(-1)-2$
$-1+2-1-2=0$
$x=-1$
$x+1=0$
So, $(x+1)$ is a factor of the polynomial $x^3+2 x^2-x-2$