105 x 106 = (100 + 5) x (100 + 6)
= (100)2 + (5 + 6) (100) + (5 x 6), using Identity IV
= 10000 + 1100 + 30
= 11130

Comparing the given expression with (x + y + z)², we find that
x = 3a, y = 4b and z = 5c.
Therefore, using Identity V, we have
(3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

9

28

Let p(x) = x² – 2x
Then p(2) = 2² – 4 = 4 – 4 = 0
and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x² – 2x.

\(35y^2+13y-12\)
\(=35y^2+28y-15y-12\)
\(=7y(5y+4)-3(5y+4)\)
\(=(5y+4)(7y-3)\)
\(\because\)  The possible expressions for the length and breadth of the rectangle are 7y-3 and 5y+4

\((-12)^3+(7)^3+(5)^3\)
\((-12)^3+(7)^3+(5)^3=3(-12)(7)(5)\)
 \(\because (-12)+(7)+(5)=0\)
\(=-1260\)

\(8a^3+b^3+12a^2b+6ab^2\)
\(=(2a)^3+(b)^3+3(2a)(b)(2a+b)\)
\(=(2a+b)^3\) | Using Identify
\(=(2a+b)(2a+b)(2a+b)\)

\(4y^2-4y+1\)
\(4y^{ 2 }-4y+1=({ 2y })^{ 2 }-2(2y)(1)+{ (1) }^{ 2 }\)
\(={ (2y-1) }^{ 2 }=(2y-1)(2y-1)\) | Using Identity II

\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​\)
\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​=\left( z+\frac { 3 }{ 2 } \right) \left( z-\frac { 3 }{ 2 } \right) ​​\)| Where \(y^2=z\)
\(={ (z) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) | Using identity III
\(={ z }^{ 2 }-\frac { 9 }{ 4 } ={ ({ y }^{ 2 }) }^{ 2 }-\frac { 9 }{ 4 } \) | Substituting the value of z
\(={ y }^{ 4 }-\frac { 9 }{ 4 } .\)

\(p(x)=2{ x }^{ 2 }+kx+\sqrt { 2 } \)
If x-1 is a factor of p(x), then p(1)=0 | By factor Theorem
\(\Rightarrow \ 2{ (1) }^{ 2 }+k(1)+\sqrt { 2 } =0\)
\(\Rightarrow \ 2+k+\sqrt { 2 } =0\)
\(\Rightarrow \ k=-(2+\sqrt { 2 } ).\)

\(p(x)={ x }^{ 3 }+{ 3x }^{ 2 }-3x+1,\ g(x)=x+2\)
g(x)=0
\(\Rightarrow x+2=0\ \Rightarrow \ x=-2\)
\(\therefore\) Zero of g(x) is -2.
Now, p(-2)
\(={ (-2) }^{ 3 }+{3 (-2) }^{ 2 }+3(-2)+1\)
\(=-8+12-6+1=-1\neq 0\)
\(\therefore\) By factor theorem, g(x) is not a factor of p(x)

Highest power of y=7

By definition

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

\(\frac{1}{y}=y^-1\) has negative exponent so, not a polynomial

Let p(x) = x³ – 23x² + 142x – 120
We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
By trial, we find that p(1) = 0. So x – 1 is a factor of p(x).
Now we see that x³ – 23x² + 142x – 120 = x³ – x² – 22x² + 22x + 120x – 120
= x²(x –1) – 22x(x – 1) + 120(x – 1) 
= (x – 1) (x² – 22x + 120) [Taking (x – 1) common]
We could have also got this by dividing p(x) by x – 1.
Now x² – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
x² – 22x + 120 = x² – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
So, x³ – 23x² – 142x – 120 = (x – 1)(x – 10)(x – 12)