Maths M.C.Q

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

$ (102)^3 = (100 + 2)^3$
$= (100)^3 + (2)^3 + 3 × 100 × 2(100 + 2)$
$= 1000000 + 8 + 60000 + 1200$
$= 1061208$

 Put $x - 3 = 0,$ then $x = 3$
Therefore, value of $x^2 - ax - 15$ at $x = 3$ is zero
$⇒ 3^2 - 3a - 15 = 0$
$⇒ -6 - 3a = 0$
$⇒ a = -2$

 $\sqrt{3}$ is a constant term, so it is a polynomial of degree $0.$

$ (249)^2 - (248)^2$
$= (249 + 248) (249 - 248) [$Using identity $a^2 - b^2 = (a + b) (a - b)]$
$= 497 × 1$
$= 497$

$ p(x) = 2x^3 - kx^2 + 3x + 10$
$x + 2 = 0 ⇒ x = -2$
By the factor theorem, we know that when $p(x)$ is divided by $(x + 2),$ the remainder is $p(-2).$
Now, $p(-2) = 2(-2)^3 + k(-2)^2+ 3(-2) + 10$
$⇒ 0 = -16 - 4k - 6 + 10$
$⇒ 0 = -12 - 4k$
$⇒ 4k = -12$
$⇒ k = -3$

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

 $(3 x-5)^3$
$=(3 x)^3-(5)^3-3 \times 3 x \times 5(3 x-5)$
$=27 x^3-125-135 x^2+225 x$
$=27 x^3-135 x^2+225 x-125$

 Let $f(x) = x^3 + 10x^2 + mx + n$
Now, $x + 2 = 0 ⇒ x = -2$
and $x - 1 = 0 ⇒ x = 1$
By factor theorem,
$f(-2) = 0$
$⇒ (-2)^3 + 10(-2)^2 + m(-2) + n$
$⇒ -8 + 40 - 2m + n = 0$
$⇒ 2m - n = 32 ...(i)$
By factor theorem,
$f(1) = 0$
$⇒ (1)^3 + 10(1)^2 + m(1) + n = 0$
$⇒ m + n = -11 ...(ii)$
Adding $(i)$ and $(ii),$ we get
$3m = 21$
$⇒ m = 7$
Substituting in $(ii),$ we get
$n = -18$

 Let $p(x) = 4x^3 + 3x^2 - 4x + k$
Now,
if $(x - 1)$ is a factor of $p(x),$ then at $x = 1, p(x) = 0$
So, $p(1) = 0$
$⇒ 4(1)^3 + 3(1)^2 - 4(1) + k = 0$
$⇒ 4 + 3 - 4 + k = 0$
$⇒ k = -3$

$\text { Given: } x+y+z=9 \text { and } x y+y z+z x=23$
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$=(x+y+y)\left[(x+y+z)^2-2 x y-2 y z-2 z x-x y-y z-z x\right]$
$=(x+y+z)\left[(x+y+z)^2-3 x y-3 y z-3 z x\right]$
$=(x+y+z)\left[(x+y+x)^2-3(x y+y z+x x)\right]$
$=(9)\left[(9)^2-3(23)\right]$
$=9 \times[(81-69)]$
$=9 \times 12$
$=108$

$ x^3 + 2x^2 - x - 2$
$= x^2 (x + 2) - 1(x + 2)$
$= (x^2 - 1) (x + 2)$
$= (x + 1) (x - 1) (x + 2)$

 Clearly, $\sqrt2\text{x}^2-\sqrt3\text{x}+6$ is a polynomial in one variable because it has only non-negative integral powers of $x.$

Given, $a+b+c=9$
Hence, $(a+b+c)^2=81$
So, $a^2+b^2+c^2+2 a b+2 b c+2 c a=81$
i.e. $a^2+b^2+c^2+2(a b+b c+c a)=81$
i.e. $a^2+b^2+c^2+2(23)=81$
i.e. $a^2+b^2+c^2=81-46=35$
Now, $a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right]$
$=(9)[35-23]$
$=9 \times 12$
$=108$
Hence, correct option is $(a).$

The given polynomial is $p(x) = x^2 + x - 6$
Putting $x = 2$ in $p(x),$ we get
$p(2) = 2^2 + 2 - 6 = 4 + 2 - 6 = 0$
Therefore, $x = 2$ is a zero of the polynomial $p(x).$
Putting $x = -3$ in $p(x),$ we get
$p(-3) = (-3)^2 - 3 - 6 = 9 - 9 = 0$
Therefore, $x = -3$ is a zero of the polynomial $p(x)$
Thus, $2$ and $-3$ are the zeros of the given polynomial $p(x).$

Let $p(x)$ be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
A polynomial consisting of one term, namely zero only, is called a zero polynomoial.
So, the zero of a zero polynomial is not defined.

$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3$
Here,$a^2-b^2+b^2-c^2+c^2-a^2=0$
Therefore,
$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a^2-b^2)(b^2-c^2)(c^2-a^2)$
${[\text { Since } x^3+y^3+z^3=3 x y z \text { if } x+y+z=0]}$
$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=$
$3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)$

$x^2 + kx -3 = (x - 3) (x + 1),$
$⇒ x^2 + kx - 3 = x^2 + (-3 + 1) x + (-3) × 1$
$⇒ x^2 + kx - 3 = x^2 = 2x - 3$
On comparing the term, we get $= -2$

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

 For making perfect square to $x^4+x^2+25$
We add $+10 x^2$ and $-10 x^2$ to it.
$=x^4+x^2+25$
$=x^4+x^2+25+10 x^2-10 x^2$
$=[x^4+10 x^2+25]-9 x^2$
$=(x^2+5)^2+(3 x)^2$
$=[(x^2+5)+3 x][(x^2+5)-3 x]$
$=(x^2+3 x+5)(x^2-3 x+5)$
Hence, correct option is $(a).$

 $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4 ($given$)$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(4)^2-2=16-2=14\ ...(1)$
Squaring equation $(1)$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2(\text{x}^2)\frac{1}{\text{x}^2}=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=194$
Hence, correct option is $(b).$

 A polynomial containing two non-zero terms is called a binomial.
Example: $3x + 4, 5y + 1, x^2 + x$

$ 2x^2 + 7x - 4$
$= 2x^2 + 8x - x - 4$
$= 2x(x + 4) -1(x + 4)$
$= (2x - 1) (x + 4)$
$2x - 1 = 0$ and $x + 4 = 0$
$\text{x}=\frac{1}{2}$ and $\text{x}-4$
Therefore, one zero of the given polynomial is $\frac{1}{2}$

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

 $\text{x}^3-\Big(\frac{1}{\text{x}^3}\Big)=110$
$\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=110+\text{3}\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)63=110+3\Big(\text{3}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)-110=0$
Let $\text{x}+\frac{1}{\text{x}}=\text{a}$
$\Rightarrow\text{a}^3-3\text{a}-110=0$
$\Rightarrow\text{a}^3-5\text{a}^2+5\text{a}^2-25\text{a}+22\text{a}-110=0$
$\Rightarrow\text{a}^2(\text{a}-5)+5\text{a}(\text{a}-5)+22(\text{a}-5)=0$
$\Rightarrow(\text{a}-5)(\text{a}^2+5\text{a}+22)=0$
$\Rightarrow\text{a}-5=0$ or $\text{a}^2+5\text{a}+22=0$ neglected
$\Rightarrow\text{a} = 5$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=5$

The degree of any constant term $5 ($say$)$
We can write $5$ as $5 \times 1 = 5x^\circ [$Since $a^\circ = 1]$
Therefore, the degree of any constant term is $0$

$(x + y - z)^2 = (x)^2 + (y)^2 + (-z)^2 + 2 × x × y + 2 × y × (-z) + 2 × (-z) ×x$
$= x^2 + y^2 + z^2 + 2xy - 2yz - 2zx$

$(x+y)^3-\left(x^3+y^3\right)$
$=x^3+y^3+3 x y(x+y)-\left(x^3+y^3\right)$
$=3 x y(x+y)$
Thus, the factors of $(x+y)^3-\left(x^3+y^3\right)$ are $3 x y$ and $(x+y)$