x(y-10)+6(y-10)
^^^that's the first step, then:
the x and 6 pair together, and one of the (y-10) leaves the expression so it becomes:
(x+6)(y-10)
this is referred to as Factoring by Grouping
If that's +14x + 3, the answer is (2x + 3)(4x + 1)
The LCM is 455x.
x3 + 2x2 - 35x = x(x2 + 2x - 35) = x(x + 7)(x - 5)
The correct answer is one of: A: (7x + 5)*(7x + 6) = 49x2 + 77x + 30 B: (7x + 5)*7x + 6 = 49x2 + 35x + 6 C: 7x + 5*7x + 6 = 42x + 6 Since there are no brackets in the question, the correct answer is C but I suspect there should be brackets but this site does not allow them.
35
(5x - 7)(7x + 5)
25x + 35x + 9 is 60x + 9, which factors to 3(20x + 3)
7(5x + 4y)
x2-35x+300 = 0 (x-15)(x-20) = 0 x = 15 or x = 20
If that's +14x + 3, the answer is (2x + 3)(4x + 1)
Oh, what a happy little math problem we have here! To factor 35x + 63y, we can first look for the greatest common factor of the coefficients, which is 7. Then we can factor out the 7 to get 7(5x + 9y). And just like that, we've created a beautiful and simplified expression!
35x + 115 = 360 35x = 360-115 35x = 245 x = 245/35 x = 7
42x+35x-35x=42x it's leave 85 so it's change into a negative 42x-85 is the answer
35x + 10
(7x-3)(2x+5) 14x2 + 29x - 15, write 29x = 35x - 6x = 14x2 - 6x + 35x - 15 = (14x2 - 6x) + (35x - 15) = 2x(7x - 3) + 5(7x - 3) = (7x-3)(2x+5)
x3 + 2x2 - 35x = x(x + 7)(x - 5)
35x*2=5x Well, I can only think of one way to solve this. Forgive if I'm wrong 'cause I'm only in 9th grade. I've tried it the other way where you solve the equation for zero, but it didn't work. So here's the only way I saw, I apologise if it is wrong. 35x*2=5x -5x......-5x 30x*2=0 .......-2-2 ..30x=-2 ...30x/30-2/30 x=-1/15