Problem : Resolve the polynomial
P
(
x
)
=
18
x
3
+
15
x
2
−
x
−
2
{\displaystyle P(x)=18x^{3}+15x^{2}-x-2}
Solution : Here, the constant term of
P
(
x
)
{\displaystyle P(x)}
−
2
{\displaystyle -2}
−
2
{\displaystyle -2}
F
{\displaystyle F}
1
=
{\displaystyle =}
Here, the leading coefficient of
P
(
x
)
{\displaystyle P(x)}
18
{\displaystyle 18}
18
{\displaystyle 18}
F
{\displaystyle F}
2
=
{\displaystyle =}
Now consider
P
(
a
)
{\displaystyle P(a)}
a
=
r
s
,
r
∈
F
{\displaystyle a={\frac {r}{s}},r\in F}
1
,
s
∈
F
{\displaystyle ,s\in F}
2
When,
a
=
1
,
P
(
1
)
=
18
+
15
−
1
−
2
≠
0
{\displaystyle a=1,P(1)=18+15-1-2\neq 0}
a
=
−
1
,
P
(
−
1
)
=
−
18
+
15
+
1
−
2
≠
0
{\displaystyle a=-1,P(-1)=-18+15+1-2\neq 0}
a
=
−
1
2
,
P
(
−
1
2
)
=
18
(
−
1
8
)
+
15
(
−
1
4
)
+
1
2
−
2
=
0
{\displaystyle a=-{\frac {1}{2}},P(-{\frac {1}{2}})=18(-{\frac {1}{8}})+15(-{\frac {1}{4}})+{\frac {1}{2}}-2=0}
Therefore,
(
x
+
1
2
)
=
1
2
(
2
x
+
1
)
{\displaystyle (x+{\frac {1}{2}})={\frac {1}{2}}(2x+1)}
P
(
x
)
{\displaystyle P(x)}
Now,
P
(
x
)
=
18
x
3
+
15
x
2
−
x
−
2
{\displaystyle P(x)=18x^{3}+15x^{2}-x-2}
=
18
x
3
+
9
x
2
+
6
x
2
+
3
x
−
4
x
−
2
{\displaystyle =18x^{3}+9x^{2}+6x^{2}+3x-4x-2}
=
9
x
2
(
2
x
+
1
)
+
3
x
(
2
x
+
1
)
−
2
(
2
x
+
1
)
{\displaystyle =9x^{2}(2x+1)+3x(2x+1)-2(2x+1)}
=
(
2
x
+
1
)
(
9
x
2
+
3
x
−
2
)
{\displaystyle =(2x+1)(9x^{2}+3x-2)}
=
(
2
x
+
1
)
(
9
x
2
+
6
x
−
3
x
−
2
)
{\displaystyle =(2x+1)(9x^{2}+6x-3x-2)}
=
(
2
x
+
1
)
(
3
x
(
3
x
+
2
)
−
1
(
3
x
+
2
)
)
{\displaystyle =(2x+1)(3x(3x+2)-1(3x+2))}
=
(
2
x
+
1
)
(
3
x
+
2
)
(
3
x
−
1
)
{\displaystyle =(2x+1)(3x+2)(3x-1)}
∴
P
(
x
)
=
(
2
x
+
1
)
(
3
x
+
2
)
(
3
x
−
1
)
{\displaystyle P(x)=(2x+1)(3x+2)(3x-1)}
Problem : Resolve the polynomial
P
(
x
)
=
−
3
x
2
−
2
x
y
+
8
y
2
+
11
x
−
8
y
−
6
{\displaystyle P(x)=-3x^{2}-2xy+8y^{2}+11x-8y-6}
Solution : Considering only the terms of
x
{\displaystyle x}
−
3
x
2
+
11
x
−
6
{\displaystyle -3x^{2}+11x-6}
−
3
x
2
+
11
x
−
6
≡
(
−
3
x
+
2
)
(
x
−
3
)
.
.
(
i
)
{\displaystyle -3x^{2}+11x-6\equiv (-3x+2)(x-3)..(i)}
In the same way, considering only the terms of
y
{\displaystyle y}
8
y
2
−
8
y
−
6
{\displaystyle 8y^{2}-8y-6}
8
y
2
−
8
y
−
6
≡
(
4
y
+
2
)
(
2
y
−
3
)
.
.
(
i
i
)
{\displaystyle 8y^{2}-8y-6\equiv (4y+2)(2y-3)..(ii)}
Combining factors of above (i) and (ii), the factors of the given polynomial can be found. But the constants
+
2
,
−
3
{\displaystyle +2,-3}
x
{\displaystyle x}
y
{\displaystyle y}
∴
P
(
x
)
=
(
−
3
x
+
4
y
+
2
)
(
x
+
2
y
−
3
)
{\displaystyle P(x)=(-3x+4y+2)(x+2y-3)}
