Please solve fast Q No. 3 , 4, 5, 6, 7 , 8 ,9 and 10.

Step-by-step explanation:

3)

Given expression is x(6x-8y)-x²(6x-5y)

=>6x²-8xy-6x³+5x²y

=> x(6x-8y-6x²+5xy)

x(6x-8y)-x²(6x-5y) = x(6x-8y-6x²+5xy)

If the given problem like this

x(6x-5y)-x²(6x-5y)

=> (6x-5y)(x-x²)

=> (6x-5y)(x)(1-x²)

=> (6x-5y)(x)(1+x)(1-x)

4)

Given expression is 9m(p-3q)-3n(p-3q)

=> (p-3q)(9m-3m)

=> (p-3q)3(3m-n)

=> 3(p-3q)(3m-n)

9m(p-3q)-3n(p-3q) = 3(p-3q)(3m-n)

5)

Given that ax+ay-bx-by

=> a(x+y)-b(x+y)

=> (x+y)(a-b)

ax+ay-bx-by = (x+y)(a-b)

6)

Given that

a²+2a+ab+2b

=> (a²+ab)+(2a+2b)

=> a(a+b)+2(a+b)

=> (a+b)(a+2)

a²+2a+ab+2b = (a+b)(a+2)

7)

Given that

2ax+bx+2ay+by

=> (2ax+2ay)+(bx+by)

=> 2a(x+y)+b(x+y)

=> (x+y)(2a+b)

2ax+bx+2ay+by = (x+y)(2a+b)

8)

Given that

xy-pq+qy-px

=> (xy+qy)-(pq+px)

=> y(x+q)-p(q+x)

=> y(x+q)-p(x+q)

=> (x+q)(y-p)

xy-pq+qy-px = (x+q)(y-p)

9)

Given that

1+p+pq+p²q

=> (1+p)+(pq+p²q)

=> 1(1+p)+pq (1+p)

=> (1+p)(1+pq)

1+p+pq+p²q = (1+p)(1+pq)

10)

Given that

2a-4b-xa+2bx

=> (2a-4b)-(xa-2bx)

=> 2(a-2b)-x(a-2b)

=> (a-2b)(2-x)

2a-4b-xa+2bx = (a-2b)(2-x)

Hope this helps!!

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.