## anonymous 3 years ago Simplify the following expression: (5x2 + 3x + 4) − (2x2 + 5x − 1). If the final answer is written in the form Ax2 + Bx + C, what is the value of A? The 2 be hide 5x is squared and so is the one by 2x

1. ryan123345

Distribute the negative

2. anonymous

So -1?

3. anonymous

I have terrible drawing skills, sorry.

4. anonymous

|dw:1363207136675:dw|

5. anonymous

So 3x - 5 - 5x + 1

6. anonymous

I meant:|dw:1363207208912:dw|

7. anonymous

Very close. What is -1 times 1?

8. anonymous

-1

9. anonymous

Right! So your equation would be 3x - 5 - 5x - 1. Can you combine the like terms?

10. anonymous

Yes. the x

11. anonymous

Lets rearrange it so its easier to work with. Remember to keep the symbol in front of each term. 3x - 5x - 5 - 1 First off, what is 3x - 5x?

12. anonymous

2x

13. anonymous

-2x-6

14. anonymous

Close, it would be a negative 2.

15. anonymous

16. anonymous

Ok

17. anonymous

So now we have: -2x - 5 - 1 -5 - 1 = ?

18. anonymous

6

19. anonymous

Negative 6. (: So our expression is: -2x - 6

20. anonymous

Ok. Thanks!

21. anonymous

You're welcome! Oh and:$\Huge \mathcal{\text{Welcome To OpenStudy} ☺}$

22. anonymous

Thanks! can you help me some more

23. anonymous

Sure thing!

24. anonymous

Simplify the following expression: (5x2 + 3x + 4) − (2x2 + 5x − 1). If the final answer is written in the form Ax2 + Bx + C, what is the value of A?

25. anonymous

The 2 be hide 5x is squared and so is the one by 2x

26. anonymous

$(5x^2 + 3x + 4) - (2x^2 + 5x - 1)$We have to distribute the -1. $(-1)\stackrel{\Huge\curvearrowright}(2x^2 + 5x - 1)$-1 * 2x^1 = -2x^2 -1 * 5x = -5x -1 * -1 = 1 Now we have: $5x^2 + 3x + 4 - 2x^2 - 5x + 1$Combine like terms. 5x^2 - 2x^2 = 3x^2 3x - 5x = -2x 4 + 1 = 5 And we end up with: $\large 3x^2 - 2x + 5$

27. anonymous

$\large Ax^2 + Bx + C$ $\large 3x^2 - 2x + 5$ So what is A?

28. anonymous

3x2

29. anonymous

Just the 3. (:

30. anonymous

Thanks! I have some more if that is ok?

31. anonymous

I've gtg right now, sorry. :/ Maybe @hartnn or @Luis_Rivera can continue to help you. I hope I helped!

32. anonymous

You did thanks!

33. hartnn

one suggestion though, make new post for new question. so that many others can help you too :)

34. anonymous

How do I do that?

35. hartnn

you can 'close' this question .

36. anonymous

Ok. Thanks

37. hartnn

good ! :) welcome ^_^