We have 50x² - 98 = 2(25x² - 49)
= 2[(5x)² - (7)²] == 2[(5x + 7)(5x - 7)] [Using a² - b² = (a + b)(a - b)]
= \(\frac { 81x^{ 3 }[50x^{ 2 }-98] }{ 27x^{ 2 }[5x+7] } \)
=\(\frac { { 81x }^{ 3 } }{ { 27x }^{ 2 } } \left[ \frac { 2(5x-7)(5x+7) }{ (5x+7) } \right] =3x[2(5x-7)]\)
= 3x\(\times \)2\(\times \)(5x- 7) = 6 x (5x - 7)

We have -45p³
= (-1) x 3 x 3 x 5 x P x P x P
9p² = 3 x 3 x p x p
∴ [- 45p³] + 9p² = [(-1) x 3 x 3 x 5 x P xP x p] \(\div \) (3 x 3 x p x p)
=\(\frac { (-1)\times 3\times 3\times 5\times P\times P\times P }{ 3\times 3\times p\times p } \)
=\(\frac { -1\times 5\times P }{ 1\times 1 } =-1\times 5p=-5p\)

We have 63(p⁴ + 5p³ - 24p²) + 9p(P + 8)
= \(\frac { 63({ p }^{ 4 }+5p^{ 3 }-24p^{ 2 } }{ 9p(p+8) } \)
= \(\frac { 63p^{ 2 }(p^{ 2 }+5p-24) }{ 9p(p+8) } \)
= \(\frac { 63p^{ 2 } }{ 9p } \)
= \(\left[ \frac { { p }^{ 2 }8p-3p-24 }{ p+8 } \right] \)
= \(7p\left[ \frac { p(p+8)-3(p+8) }{ (p+8) } \right] \)
= \(7p\left[ \frac { (p+8)(p-3) }{ (p+8) } \right] =7p(p-3)\) 

By regrouping, we have
x - 9 + 9zy - xyz = x - 9 - xyz + 9zy
= 1(x - 9) - yz(x - 9) = (x - 9)(1 - yz)

we have ax³y²=a\(\times \)x\(\times \)x\(\times \)x\(\times \)y\(\times \)y
bx²y³=b\(\times \)x\(\times \)x\(\times \)y\(\times \)y\(\times \)y
cx²y²z=c\(\times \)x\(\times \)x\(\times \) y\(\times \)y\(\times \)z
Obviously, x²y² is a common factor.
ஃ we get ax³y²=x²y²\(\times \)a\(\times \)
bx²y³=x²y²\(\times \)b\(\times \)y
cx²y²z=x²y²\(\times \)c\(\times \)z
\(\Rightarrow \) ax³y²+bx²y³+cx²y²z
= [(x²y²\(\times \)a\(\times \)x)+(x²\(\times \)y²\(\times \)b\(\times \)y) + x²y²\(\times \) c\(\times \) z)]
= x²y²[(a\(\times \)x)+(b\(\times \)y)+(c\(\times \)z)]
= x²y²(ax + by + cz)

We have 27x³ = 3 x 3 x 3 \(\times \)x\(\times \)x\(\times \)x
21x² = 3\(\times \)7\(\times \)x\(\times \)x
15x⁴ = 3 x 5 \(\times \) x \(\times \) x \(\times \) x \(\times \) x
We get common factors as 3, x and x.
i.e., 3 \(\times \) x \(\times \)x or 3x²
∴27x³ =: 3x² x 3 \(\times \) x \(\times \) 3 = 3x² \(\times \)9x
21x²=\({ 3x }^{ 2 }\times 7={ 3x }^{ 2 }\times 7\)
\({ 15x }^{ 4 }=3x^{ 2 }\times 5x\times x={ 3x }^{ 2 }\times { 5x }^{ 2 }\)
\(\Rightarrow { 27x }^{ 3 }-{ 21x }^{ 2 }+{ 15x }^{ 4 }\)
\(\Rightarrow { (3x }^{ 2 }\times 9x)-({ 3x }^{ 2 }\times 7)+{ (3x }^{ 2 }\times { 5x }^{ 2 })\)
\({ 3x }^{ 2 }[9x-7+{ 5x }^{ 2 }]\)
 

Given mathematical statement is incorrect.
Hence, correct statement is 2x + 3y = 2x + 3y because in both terms, variables are different and we can add only like terms

It is given that, the expression (x² + 19x - 20) shows the area of rectangle. To find the possible length and breadth of this rectangle. We have to factorise (x² + 19x - 20).
Since, -1 and 20 are integers such that
-1 x 20 = - 20 and 20 + (-1) = 19
Putting these values in given expression
x² + (20 -1) x - 20 = x² + 20x - x - 20
= x (x + 20) -1 (x + 20)
=(x + 20) (x -1)
Hence, possible length and breadth of this rectangle will be (x + 20) and (x -1).

Here, 5P²-25P+20 = 5(p²-5P + 4)
= 5[p² +(-4-1)p+4]
= 5(p²-4p-P+ 4)    [∵ ab = 4 and a + b= -5, ஃ a= -4, b= -1]
= 5[p(p-4)-1(p-4)]
= 5(p-4)(p-1)
Now, (5p²-25p + 20) ÷ (p-1)
=\(\frac{5p^{2}-25p+20}{(p-1)}\)
=\(\frac{5(p-4)(p-1)}{(p-1)}=5(p-4)\)

p⁴ -81=(p²)² -(9)²
On comparing with a² - b², we get a = p² and b = 9
p⁴ -81 = (p² + 9)(p² -9)
[\(\therefore\) (a² -b²) = (a + b)(a - b)]
= (p²- 9) [(p)² - (3)²]
= (p²+ 9) (p + 3 (p - 3)
[\(\because\) a²-b2 =(a+b)(a-b) and (p² + 9) cannot be factorised further]

a⁴-b⁴ = (a²)²-(b²)²
Oncomparing with a² - b², we get a = a² and b = b²
\(\therefore\)  a⁴ -b⁴ = (a² +b²)(a² -b²)
= (a² + b² )(a + b)(a - b) [\(\because\) a²-b2 = (a + b)(a - b)] and a² + b² cannot be factorised further

25m² + 30m + 9 = (5m)² + 30m + (3)²
= (5m)² + 2(5)m (3) + (3)²
This expression is of the form a² + 2ab + b².
On comparing, we get a = 5m and b = 3
Since, a² + 2ab + b² = (a + b)²
\(\therefore\) 25m² + 30m + 9 = (5m + 3)2 or (5m + 3)(5m + 3)

p² -l0p + 25 = (p)² -l0p + (5f = p² - 2(5)p + 5²
This expression is of the form a 2 - 2ab + b² .
On comparing, we get a = p and b = 5
Since, a² - 2ab + b² = (a - b)²
\(\therefore\) p² -10p+25=(p-5)² or (p-5)p-5²

we have  -16z = (-1) x 2 x 2 x 2 x 2 x z
and  20z³ = 2 x 2 x 5 x z x z X z
The two terms have 2, 2 and z as common factors
Therefore, -16z + 20z³ = (-1) x 2 x 2 x 2 x 2 x z x 2 x 2 x 2 x 5 x z x z x z
= 2 x 2 x z [(-1) x 2 x 2 + 5 x z x z] [combining the terms]
= 4z (-4 + 5z²)

we have 7a² = 7 x a x a and 14a = 2 x 7 x a
the two terms have 7 and as common factors
therefore. 7a² + 14a = 7 x a  x a + 2 x 7 x a
= 7 x a (a+2)  [combining the terms]
= 7 a (a + 2)
It is the required factor form