MATHS M.C.Q

$4 a^2+b^2+4 a b+8 a+4 b+4$
$=4 a^2+b^2+4+4 a b+4 b+8 a$
$=(2 a)^2+b^2+2^2+2 \times 2 a \times b+2 \times b \times 2+2 \times 2 a \times 2$
$=(2 a+b+2)^2$

$\frac{a}{b}+\frac{b}{a}=-1$
$\Rightarrow \frac{a^2+b^2}{a b}=-1$
$\Rightarrow a^2+b^2=-a b$
$\Rightarrow a^2+b^2+a b=0$
Thus, we have:
$\left(a^3-b^3\right)=(a-b)\left(a^2+b^2+a b\right)$
$=(a-b) \times 0$
$=0$

$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$=(x+y+z)\left[(x+y+z)^2-3(x y+y z+z x)\right]$
$=9 \times(81-3 \times 23)$
$=9 \times 12$
$=108$

$x^2-4 x-21$
$x^2-7 x+3 x-21$
$=x(x-7)+3(x-7)$
$=(x-7)(x+3)$

$(x+3)^3$
$=x^3+3^3+9 x(x+3)$
$=x^3+27+9 x^2+27 x$
So, the coefficient of $x$ in $(x+3)^3$ is $27$ .

$6 x^2+17 x+5$
$=6 x^2+15 x+2 x+5$
$=3 x(2 x+5)+1(2 x+5)$
$=(2 x+5)(3 x+1)$

$a+b+c=0 \Rightarrow a^3+b^3+c^3=3 a b c$
Thus, we have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$

$(x+1)$ is a factor of $2 x^2+k x$
So, $-1$ is a zero of $2 x^2+k x$
Thus, we have:
$2 \times(-1)^2+k \times(-1)=0$
$\Rightarrow 2-\mathrm{k}=0$
$\Rightarrow \mathrm{k}=2$

$(249)^2-(248)^2$
We know
$a^2-b^2=(a+b)(a-b)$
So,
$(249)^2-(248)^2$
$(249-248)(249+248)$
$=497$

$207 \times 193$
$=(200+7)(200-7)$
$=(200)^2-(7)^2$
$=40000-49$
$=39951$

Let:
$f(x)=x^3-2 x^2+x+2$
By the factor theorem, $(x+1)$ will be a factor of $f(x)$ if $f(-1)=0$.
We have:
$f(-1)=(-1)^3-2 \times(-1)^2+(-1)+2$
$=-1-2-1+2$
$=-2 \neq 0$
Hence, $(x+1)$ is not a factor of $f(x)=x^3-2 x^2+x+2$.
Now,
Let:
$f(x)=x^3+2 x^2+x-2$
By the factor theorem, $(x+1)$ will be a factor of $f(x)$ if $f(-1)=0$.
We have:
$f(-1)=(-1)^3+2 \times(-1)^2+(-1)-2$
$=-1+2-1-2$
$=-2 \neq 0$
Hence, $(x+1)$ is not factor of $f(x)=x^3+2 x^2+x-2$.
Now,
Let:
$f(x)=x^3+2 x^2-x-2$
By the factor theorem, $(x+1)$ will be a factor of $f(x)$ if $f(-1)=0$.
We have:
$f(-1)=(-1)^3+2 \times(-1)^2-(-1)-2$
$=-1+2+1-2$
$=0$
Hence, $(x+1)$ is a factor of $f(x)=x^3+2 x^2-x-2$.

$a+b+c=0$
$\Rightarrow a+b=-c$
$\Rightarrow(a+b)^3=(-c)^3$
$\Rightarrow a^3+b^3+3 a b(a+b)=-c^3$
$\Rightarrow a^3+b^3+3 a b(-c)=-c^3$
$\Rightarrow a^3+b^3+c^3=3 a b c$

$\frac{x}{y}+\frac{y}{x}=-1$
$\Rightarrow \frac{x^2+y^2}{x y}=-1$
$\Rightarrow x^2+y^2=-x y$
$\Rightarrow x^2+y^2+x y=0$
Thus, we have:
$\left(x^3-y^3\right)=(x-y)\left(x^2+y^2+x y\right)$
$=(x-y) \times 0$
$=0$

$\left(25 x^2-1\right)+(1+5 x)^2$
$=(5 x-1)(5 x+1)+(1+5 x)^2$
$=(5 x+1)[(5 x-1)+(1+5 x)]$
$=(5 x+1)(10 x)$
So, the factors of $\left(25 x^2-1\right)+(1+5 x)^2$ are $(5 x+1)$ and $10 x$

$305 \times 308=(300+5)(300+8)$
$=(300)^2+300 \times(5+8)+5 \times 8$
$=90000+3900+40$
$=93940$

$104 \times 96=(100+4)(100-4)$
$=100^2-4^2$
$=(10000-16)$
$=9984$

Let:
$p(x)=x^3+10 x^2+m x+n$
Now,
$x+2=0 \Rightarrow x=-2$
$(x+2)$ is a factor of $p(x)$.
So, we have $p(-2)^2+m \times(-2)+n=0$
$\Rightarrow(-2)^3+10 \times(-2)^2+m \times(-2)+n=0$
$\Rightarrow-8+40-2 m+n=0$
$\Rightarrow 32-2 m+n=0$
$\Rightarrow 2 m-n=32 \ldots(i)$
Now,
$x-1=0 \Rightarrow x=1$
Also,
( $\mathrm{x}-1$ ) is a factor of $\mathrm{p}(\mathrm{x})$
We have:
$p(1)=0$
$\Rightarrow 1^3+10 \times 1^2+m \times 1+n=0$
$\Rightarrow 1+10+m+n=0$
$\Rightarrow 11+m+n=0$
$\Rightarrow m+n=-11 \ldots \text { (ii) }$
From $(i)$ and $(ii)$,
We get:
$3 m=21 \Rightarrow m=7$
By substituting the value of $m$ in $(i)$, we get $n=-18$
$\therefore m=7 \text { and } n=-18$

$(x+5)$ is a factor of $p(x)=x^3-20 x+5 k$
$\therefore p(-5)=0$
$\Rightarrow(-5)^3-20 \times(-5)+5 k=0$
$\Rightarrow-125+100+5 k=0$
$\Rightarrow 5 k=25$
$\Rightarrow k=5$