Maths M.C.Q

$x+1$ is a factor of $p(x)=2 x^2+k x$
Then, $\mathrm{p}(-1)=0$
i.e. $2(-1)^2+k(-1)=0$
$2-k=0$
$k=2$

Let $f(x)=p x^2+5 x+r$
Now,
If $x-2$ and $x-\frac{1}{2}$ are factors of $f(x)$.
Then at $x=2$ and $x-\frac{1}{2}, f(x)=0$.
So, $f(2)=0, f\left(\frac{1}{2}\right)=0$
$\Rightarrow \mathrm{p}(2)^2+5(2)+\mathrm{r}=0$
And, $\mathrm{p}\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\right)+\mathrm{r}=0$
$\Rightarrow 4 p+r+10=0 \ldots(1)$
And $4 r+p+10=0 \ldots(2)$
Subtracting equation $(2)$ from $(1)$, we have
$3 p-3 r=0$
$\Rightarrow p=r$

Since, $p(x)=x^3+6 x^2+4 x+k$ is exactly divisible by $x+2$
$(x+2)$ is a factor of $p(x)$.
So, $p(-2)=0$
i.e. $(-2)^3+6(-2)^2+4(-2)+k=0$
$-8+24-8+k=0$
$24-16+k=0$
$8+k=0$
$k=-8$

Let $p(x)=x^3-3 x^2 a+2 a^2 x+b$
$(x-a)$ is a factor of $p(x)$.
So,
$p(a)=0$
$a^3-3 a^2 a+2 a^2 a+b=0$
$a^3-3 a^3+2 a^3+b=0$
$3 a^3-3 a^3+b=0$
$b=0$

Let $p(x)=x^2+3 a x-2 a$ be the given polynomial.
$x-2$ is a factor of $p(x)$.
Thus,
$p(2)=0$
$(2)^2+3 a \times 2-2 a=0$
$4+4 a=0$
$a=-1$

If $x-1$ is a factor of $f(x)$ then definitely $f(1)=0$
And,
$x-1$ is not a factor of $g(x)$, then $g(1) \neq 0$.
So, at $x=1$
$a.$ $f(1) g(1)=0 \times g(1)=0$
$b.$ $-f(1)+g(1)=0+g(1)=g(1) \neq 0$
$c.$ $f(1)-g(1)=0-g(1)=-g(1) \neq 0$
$d.$ $\{f(1)+g(1)\} g(1)=\{0+g(1)\} g(1)=\{g(1)\}^2 \neq 0$
So, at $x=1$ only, $f(x) g(x)=0$
Thus, $(x-1)$ is factor of $f(x) g(x)$ too.

If $x+1$ is a factor of $x^n+1$
then, at $x=-1, x^n+1=0$
$(-1)^n+1=0$
$(-1)^n=-1$
$(-1)^n$ will be equal to -1 if and only if $n$ is an odd integer.
If $n$ is even, then $(-1)^n=1$
So, $n$ should be an odd integer.

Let $\mathrm{p}(\mathrm{x})=\mathrm{x}^{140}+2 \mathrm{x}^{151}+\mathrm{k}$
Since $p(x)$ is divisible by $(x+1)$,
$(x+1)$ is a factor of $p(x)$.
So,
$p(-1)=0$
$(-1)^{140}+2(-1)^{151}+k=0$
$1+2(-1)+k=0$
$1-2+k=0$
$k-1=0$
$k=1$

$x^4+x^2-20$
$=x^4+5 x^2-4 x^2-20$
$=x^2\left(x^2+5\right)-4\left(x^2+5\right)$
$=\left(x^2+5\right)\left(x^2-4\right)$
So, other factor is $\mathrm{x}^2-4$

If $x^2-1$ is factor of $p(x)=a x^4+b x^3+c x^2+d x+e$.
Then $(x-1)$ and $(x+1)$ will also be factors of $p(x)$.
Because $x^2-1=(x-1)(x+1)$
Then, at $x=1$ and $x=-1, p(x)=0$
$\Rightarrow p(1)=0 \text { and } p(-1)=0$
$\Rightarrow a+b+c+d+e=0 \ldots \text { (1) }$
And
$\Rightarrow a-b+c-d+e=0 \ldots \text { (2) }$
Adding equations $(1)$ and $(2)$.
$2 a+2 c+2 e=0$
$\Rightarrow a+c+e=0 \ldots \text { (3) }$
Substracting equation $(2)$ from $(1)$
$2 b+2 d=0$
$\Rightarrow b+d=0 \ldots \text { (4) }$
From equations $(3)$ and $(4)$, we get
$a+c+e=b+d$

Let $p(x)=3 x^3+8 x^2+8 x+3+5 k$ and $q(x)=x^2+x+1$
Now,
If $q(x)$ is a factor of $p(x)$, then arranging $p(x)$ in order to have $q(x)$ in common,
$p(x)=3 x^3+3 x^2+3 x+5 x^2+5 x+3+2-2+5 k[\text { Adding }+2,-2 \text { in } p(x)]$
$=3 x\left(x^2+x+1\right)+5\left(x^2+x+1\right)+5 k-2$
$p(x)=\left(x^2+x+1\right)(3 x+5)+5 k-2 \ldots(1)$
From equation $(1)$, we can see if we divide $p(x)$ by $q(x)$, then quotient will be $(3 x+5)$ and remainder will be ( $5 k-2)$
But $q(x)$ is a factor of $p(x)$.
So, remainder $=0$
$\Rightarrow 5 \mathrm{k}-2=0$
$\Rightarrow \mathrm{k}=\frac{2}{5}$

$(3 x-1)^7=a_7 x^7+a_6 x^6+\ldots+a_1 x+a_0 \ldots(1)$
Putting $x=1$ in equation $(1)$, we have
$[3(1)-1]^7=a_7+a_6+\ldots . .+a_1+a_0$
So,
$a_7+a_6+a_5+\ldots . .+a_1+a_0=2^7=128$

If $x+2$ is a factor of $x^2+m x+14$
then at $x=-2$,
$x^2+m x+14=0$
i.e. $(-2)^2+m(-2)+14=0$
$4-2 m+14=0$
$2 m=18$
$m=9$

$x+a$ is a factor of polynomial $f(x)=x^4-a^2 x^2+3 x-6 a$,
Then at $x=-a, p(x)=0$
$\Rightarrow(-a)^4-a^2(-a)^2+3(-a)-6 a=0$
$\Rightarrow a^4-a^4-3 a-6 a=0$
$\Rightarrow-9 a=0$
$\Rightarrow a=0$