Question

60x^2-68x+8

Answer 1

4(15x^2-17x+2)
4(15x^2-15x-2x+2)
4(15x(x-1)-2(x-1))
4(15x-2)(x-1))

Answer 2

thank you do you mind if u do 7x^3-14x^2-3x+6

Answer 3

7x^2(x-2) - 3(x-2)
(7x^2 - 3)(x-2)

Answer 4

thanks again but if u really do not mind can u please do 30x^2+95xy+50y^2 and2x^2+xy-3y^2 and x^3-26x^2+25 thanks a million u r a lifesaver i have a teacher who can not teach and i do not learn anything. Again thank you

Answer 5

I figured that maybe you would have understood the method by now

Answer 6

I only understand a few but some are really diffucult that I have no clue how to start

Answer 7

The first step is to always see if you can find something common to each term. If so, factor it out. Make sure you look at the coefficients and the variables

Answer 8

Once you do that, what's left inside the parentheses may be in a form where you can factor it.

Answer 9

soo can u show me step by step how to solve 30x^2+95xy+50y^2

Answer 10

The first step is to find the highest factor common to each term...(coefficients and variables). In this case, only the coefficients have a common factor. And that factor is 5. So we factor it out:

5(6x^2 + 19xy + 10y^2)

Now we're left with something simpler inside the parentheses. The part in the parentheses above is all that is left that we need to factor.

So, The next step is to find a number that multiplies to get 60 but add to get 19. By doing this, it will allow us to split the middle term into two factors so that we can factor by grouping.

By trial and error, those factors are 4 and 15.

4 + 15 = 19
4*15 = 60

So now, we re-write the middle term below like so:

5(6x^2+4xy + 15xy + 10y^2)

Now, we have to factor by grouping. To do that, we simply factor out what's common to the first two terms inside the parentheses: 6x^2 and 4xy, then what's common to the last two terms inside the parentheses: 15xy and 10y^2

For the first two terms, 2x is common
For the last two terms, 5y is common:

5(2x(3x+ 2y)+5y(3x+2y))

Now, once again, paying attention only to what's inside the parentheses, we see that 3x+2y is common to two inner terms inside the parentheses. Notice that 2x+5y are outside the (3x+2y). So when we factor out (3x+2y) we get this:

5((3x+2y)(2x+5y))

And that's the final answer. Sorry it took so long to explain.

Answer 11

thats ok that really helped me and i think i am starting to understand thanks again