Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
4(15x^2-17x+2) 4(15x^2-15x-2x+2) 4(15x(x-1)-2(x-1)) 4(15x-2)(x-1))
thank you do you mind if u do 7x^3-14x^2-3x+6
7x^2(x-2) - 3(x-2) (7x^2 - 3)(x-2)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
I figured that maybe you would have understood the method by now
I only understand a few but some are really diffucult that I have no clue how to start
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
Once you do that, what's left inside the parentheses may be in a form where you can factor it.
soo can u show me step by step how to solve 30x^2+95xy+50y^2
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.
thats ok that really helped me and i think i am starting to understand thanks again

Not the answer you are looking for?

Search for more explanations.

Ask your own question