Question

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Answer 1

see explanation

Explanation:

Using the rules of exponents

`(a^(m) )/(a^(n) )` ⇔
`a^((m-n))` ( m > n)

`(a^(m) )/(a^(n) )` ⇒
`(1)/(a^((n-m)) )` ( n > m )

Given

`(21x^6y^5)/(14x^2y^9)`

=
`(21)/(14)` ×
`x^((6-2))` ×
`(1)/(y^((9-5)) )`

=
`(3)/(2)` ×
`x^(4)` ×
`(1)/(y^(4) )`

=
`(3x^4)/(2y^4)` ← with positive exponents

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(a)

x² + 2x ← factor out x from each term

= x(x + 2)

(b)

2x² - 6x ← factor out 2x from each term

= 2x(x - 3)

(c)

15x - 10x³ ← factor out 5x from each term

= 5x(3 - 2x²)

(d)

9x² + 3x³ ← factor out 3x² from each term

= 3x²(3 + x)

Answer 2

Simplify ans: 3 x^2 y^-4

a no = x( x+2)

b no = 2x(x- 3)

c no = 5x(3- 2x^2)

d no.= 3x^2( 3+ x)

Hope this helps you