x = 1 
y = x² -5x + 6
y = 1 -5 + 6 = 2
x = 2
y = 4 -10 + 6 = 0
x = 0 
y = 6
x = 3
y = 9 -15 + 6 = 12
x = -1 
y = 1 + 5 + 6 = 12
x = 4
y = 16 - 20 + 6 = 2

(1, 2), (0, 6), (-1, 12)
P(x) = (x²-5x + 6) (x²-5x+16)
= x⁴- 5x³+16x²-5x+25x²- 80x + 6x²- 30x + 96 = 0
x⁴-10x³+ 47x² -110x + 96 = 0
It has two sign changes
\(\therefore\) it has two positive real roots
P(-x) = x⁴-10x³+ 47x² -110x + 96
It has no sign changes, no negative real roots
y = x²- 5x + 16

x	0	1	-1	2	4
y	16	12	23	10	12

 

This equation is Type- 2 even degree reciprocal equation. Hence, it can be rewritten as
\(6({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } )-5(x+\frac { 1 }{ x } )-38=0...(1)\)
Put \(x+\frac { 1 }{ x } =y\Rightarrow \left( x+\frac { 1 }{ x } \right) ^{ 2 }={ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ y }^{ 2 }\Rightarrow { x }^{ 2 }+\frac { 1 }{ x } ={ y }^{ 2 }-2\)
∴ (1)becomes as,

\(6{ (y }^{ 2 }-2)-5y-38=0\)
\(\Rightarrow 6{ y }^{ 2 }-12-5y-38=0\)
\(\Rightarrow 6{ y }^{ 2 }-5y-50=0\)
\(\Rightarrow (3y-10)(2y+5)=0\)
\(\Rightarrow y=\frac { 10 }{ 3 } ,\frac { -5 }{ 2 } \)
Case (i) when \(y=\frac { +10 }{ 3 } \)
\(x+\frac { 1 }{ x } =\frac { +10 }{ 3 } \)
\(\frac { { x }^{ 2 }+1 }{ x } =\frac { +10 }{ 3 } \Rightarrow { 3x }^{ 2 }+3=10x\)
\({ 3x }^{ 2 }-10x+3=0\)
\((x-3)(x-\frac { 1 }{ 3 } )=0\Rightarrow x=3,\frac { 1 }{ 3 } \)
Case (ii) when \(y=-\frac { 5 }{ 2 } \Rightarrow x+\frac { 1 }{ x } =-\frac { 5 }{ 2 } \)
\(\Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { -5 }{ 2 } \)
\(\Rightarrow { 2x }^{ 2 }+2+5x=0\)
\(\Rightarrow { 2x }^{ 2 }+5x+2=0\)
\(\Rightarrow (x+2)(2x+1)=0\)
\(\Rightarrow x=-2,\frac { -1 }{ 2 } \)
\(\therefore\) The roots are \(3,\frac { 1 }{ 3 } ,-2,\frac { -1 }{ 2 } \)

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as
x²\(\left[ \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -10 \left( x+\frac { 1 }{ x } \right) +26 \right] \)= 0 Since x \(\neq\) 0, we get \(\left(x^{2}+\frac{1}{x^{2}}\right)-10\left(x+\frac{1}{x}\right)+26=0\)
Let y = \(\left( x+\frac { 1 }{ x } \right) \)
[(y²-2)-10y+26] = 0 ⇒ (y²-10y+24) = 0 ⇒ (y-6)(y-4) = 0 ⇒ y = 6 or y = 4
Case (i)
y = 6 ⇒ x +\(\frac{1}{x}\) = 6 ⇒ x = 3+2\(\sqrt{2}\), x = 3 - 2\(\sqrt{2}\)
Case (ii)
y = 4 ⇒ x = 2+\(\sqrt{3}\), x = 2-\(\sqrt{3}\).
Hence, the roots are \(3 \pm 2 \sqrt{2}, 2 \pm \sqrt{3}\)

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

(i) 
Rearrange the terms as,
(x-5) (x+4) (x-7) (x+6) = 504
{-2, 3, -7, 8}
⇒ (x² -x - 20) (x² -x - 42) = 504
Put x²- x = y
⇒ (y-20)(y-42) = 504
⇒ y²-62y+840-504 = 0
⇒ y²-62y+336 = 0
⇒ (y - 56) (y - 6) = 0
⇒ y = 56, 6
Case (i)
When y = 56, x² - x = 56

⇒ x² - x - 56 = 0
⇒ (x - 8)(x + 7) = 0
⇒ x = 8, -7
Case (ii) 
When y = 6,
x²- x = 6
x²- x - 6 = 0
⇒ (x - 3)(x + 2) = 0
⇒ x = 3, - 2

Hence the roots are -2, 3, 8, -7