solve by factoring?

2x^3-14x^2+20x=0
2x(x^2 - 7x + 10)=0
2x(x - 5)(x - 2) = 0
x = 0 or x = 5 or x = 2

2x^3-14x^2+20x=0

First factor out a 2x.

2x(x^2-7x+10)=0

Now find two numbers that multiply to equal 10 and add to equal -7.

2x(x-2)(x-5)=0

Now set each factor equal to zero.

2x=0, x=0

x-2=0, x=2

x-5=0, x=5

2x^3 - 14x^2 + 20x = 0
2(x^3 - 7x^2 + 10x) = 0
2x(x^2 - 7x + 10) = 0
2x(x - 5)(x - 2) = 0

2x = 0
x = 0/2
x = 0

x - 5 = 0
x = 5

x - 2 = 0
x = 2

∴ x = 0 , 5 , 2

you can start by factoring out a 2x, so...
2x[x^2-7x+10=0]

and most 2nd degree trinomials factor into two first degree binomials.

And this one does too....

So
2x(x-5)(x-2)=0

From here you use the zero product principal.
Either 2x, x-5, or x-2 have to equal 0, so your going to have to have three solutions.

First, 2x=0-->x=0
Next x-5=0--> x=5
and x-2=0--> x=2

And there are your three solutions.
x=0, 2, or 5

First see if there is a greatest common factor that can be factored out of the polynomial. In this case, there is one, which is 2x. The polynomial factors into 2x(x^2 - 7x + 10).

Now see if the polynomial portion of this can be factored into two binomials, looking for numbers that would make the binomials multiply out to the polynomial using FOIL. In this case, (x^2)-7x+10 = (x-2)(x-5).

So now you have 2x(x-2)(x-5)=0.
Any of these factors -- 2x, x-2, x-5 -- being equal to 0 would make the left side of the equation be 0 and therefore make the equation true. So take each of them and set it equal to 0 and solve to get each solution. There would be three solutions.
-- 2x = 0 ---> x=0
-- x-2=0 ---> x=2
-- x-5=0 ---> x=5

So the solutions are 0, 2, and 5.

2x^3-14x^2+20x=0

you can see 2x is a common factor for all the tems so:
2x(x^2-7x+10)=0

lets factor the inside ( ) number, remember that you must find 2 numbers that multiplied the answer is the independetn term (10 in this case) and added the x term (-7)

(x^2-7x+10)=(x-2)(x-5)

finally

x(x-2)(x-5)=0

you know anything times zero is zero, so at least 1 of the 3 terms is zero,
so your 3 answers for x are

x=0
(x-2)=0.... so x=2
(x-5)=0 ....so x=5

good luck!

Okay, to solve by factor, please do the following:

2x^3-14x^2+20x=0

You will need to find a factor that is common amongst the three terms. In this case, the common factor will be 2x. Note that I factored out a variable and a number. So, once you factor out 2x from each of the terms, you are left with:

2x^3-14x^2+20x=0
2x(x^2 - 7x + 10) = 0

You should now treat the left side of the equation as two separate parts:

2x = 0

and

x^2 - 7x + 10 = 0

The reason why you do so is because either of the two can be equal to 0, or that they can both be equal to 0. You have to cover all possibilities, and thus you have to equate both to 0.

So now, you can solve for each of the new equations:

2x = 0
x = 0

x^2 - 7x + 10 = 0
(x -2)(x-5) = 0

x - 2 = 0
x = 2

x - 5 = 0
x = 5

Therefore, looking at all your solved equations, x can take on three values: 0, 2, 5.

Hope this helped.

2x^3-14x^2+20x=0
Factor out a 2x

2x(x^2-7x+10)
2x(x-5)(x-2)
so x=0,x=5,and x=2 are all possible solutions for x.

2x (x² - 7x + 10) = 0
2x (x - 5)(x - 2) = 0
x = 0 , x = 5 , x = 2