\(103\times 107\)
103x107=(100+3) x (100+7)
=(100)²+(3+7)(100+(3)(7))  |Using Identity IV
=10000+1000+21=11021.
Aliter
103 x 107= (100+3) x (100+7)
=(105)^2-(2)²  | Using Identity III
=(100+5)² -4
=(100)² +2(100)(5)+(5)²-4 |Using Identity I
=10000+1000+25-4=11021.

\(p(x)={ x }^{ 3 }-{ 4x }^{ 2 }+x+6, g(x)=x-3\)
g(x) = 0
\(\Rightarrow x-3=0\ \Rightarrow \ x=3\)
\(\therefore\) Zero of g(x) is 3
Now, p(3),
\({ =(3) }^{ 3 }-{ 4(3) }^{ 2 }+3+6\)
\(\\ =27-36+3+6=0\)
\(\therefore\) By factor theorem, g(x) is a factor of p(x)

\(p(-1)=(-1+1)(-1-2)\)
\(\\ =(0)(-3)=0\)
\(\\ \ p(2)=(2+1)(2-2)=(3)(0)=0\)
\(\therefore -1,2\) are zeros of p(x)

Let f (x) = 5x - 4x² + 3
Value of f(x) at x = -1
= f (-1) = 5 (-1) - 4 (-1)² + 3
= - 5 - 4 + 3 = - 6

Coefficient of x² = -1

4x² + \(\frac{1}{4 \mathrm{x}^{2}}\) + 2 = \(\left(2 x+\frac{1}{2 x}\right)^{2}\)and 9y² = (3y)²
∴ 4x² + \(\frac{1}{4 \mathrm{x}^{2}}\) +2 - 9y² =  \(\left(2 x+\frac{1}{2 x}\right)^{2}\) - (3y)²
(2x +\(\frac{1}{2 x}\) + 3y)(2x + \(\frac{1}{2 x}\) - 3y)

Here p(x) = x² + 4x - kx + 2
If p(x) is exactly divisible by x - 2, then p(2) = 0
i.e. (2)² + 4(2) - k(2) + 2 = 0
⇒ 4 + 8 - 2k + 2 = 0
⇒ 14 - 2k = 0
⇒ 2k = 14
k = \(\frac{14}{2}\) =7
Thus, the required value of k is 7.

∵ x + y + z = 0 ⇒ x³ + y³ + z³ = 3xyz
∴ a + b + 2 = 0 ⇒ (a)³ + (b)³ + (2)³ = 3(a x b x 2) = 6ab
⇒ The value of a³ + b³ + 8 is 6ab

5

370

\(-\frac{7}{48}\)

a = \(\frac{15}{7}\),
b = \(\frac{8}{7}\)

(y-2)(y-3)

4, 67

28

monomial

Let  \(p(z)={ bz }^{ 3 }+{ 4z }^{ 2 }+3z-4\)
and  \(p(z)={ z }^{ 3 }-4z+b\)
By remainder theorem,
remainder when p(z) is divided by z-3
\(=p(3) \ \ \ \ |z-3=0\Rightarrow z=3\) 
\(={ b(3) }^{ 3 }+{ 4(3) }^{ 2 }+3(3)-4\) 
\(=27b+36++9-4\) 
\( 27b+41\ \ \ \ .........(1)\)
and remainder when q(z) is divided by z-3
\(=q(3) \ \ \ \ \ \ \ |z-3=0\Rightarrow z=3\) 
\(={ (3) }^{ 3 }-4(3)+b=27-12+b\) 
\(=b+15\ \ \ ............(2)\)
According to the question,
\(27b+41=b+15\) 
\(\Rightarrow 26b=-26\) 
\(\Rightarrow b=-1\)

We have \(g(x)={ 3x }^{ 2 }-2\)
\(g\left( \frac { 2 }{ \sqrt { 3 } } \right) =3{ \left( \frac { 2 }{ \sqrt { 3 } } \right) }^{ 2 }-2=2\neq 0\)
\(\therefore \ x=\frac { 2 }{ \sqrt { 3 } } \)is not a zero of g(x)
\(g\left( -\frac { 2 }{ \sqrt { 3 } } \right) =3{ \left( -\frac { 2 }{ \sqrt { 3 } } \right) }^{ 2 }-2=2\neq 0\)
\(\therefore x=-\frac { 2 }{ \sqrt { 3 } } \) is not a zero of g(x)

\(p(x)=(p-x)^3+14=(17-x)^3+14\)
\(=(17)^3-x^3-3(17)^2(x)+3.17.x^2+14\)
\(\because\)  Degree=3

Highest power of x=5

\((1+3x)^3\) \(=1+27x^3+9x+27x^2\) It has 4 terms

Fractional power of x in (b),
Two variables in (c),
Negative power of x in (d)