answersLogoWhite

0


Best Answer

We can combine the like terms. So the b2 can be combined with the 2b2 to give 3b2. Likewise the 3b plus the -5b gives -2b.

Therefore, b2 + 3b - 5b + 2b2 = 3b2 - 2b.

User Avatar

Wiki User

13y ago
This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: What is b2 plus 3b-5b plus 2b2?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

What is 24a plus b2 plus 3a plus 2b2?

24a + b2 + 3a + 2b2= 27a + 3b2


B to the second plus b to the second?

b2 + b2 = 2b2 (when terms are alike, just add them up)


Find the illegal values of b in the fraction 2b2 plus 3b-10 over b2-2b-8?

2b2 + 8 para b = -3


What is b squared plus ab minus two minus two b squared plus two ab?

b2 + ab - 2 - 2b2 + 2ab = -b2 + ab - 2 which cannot be simplified further.


Factor b3 - 5b2 plus 12?

b3 - 5b2 + 12 = (b - 2)(b2 - 3b - 6)Check:(b - 2)(b2 - 3b - 6)= b(b2 - 3b - 6) - 2(b2 - 3b - 6)= b3 - 3b2 - 6b - 2b2 + 6b + 12= b3 - 5b2 + 12


How do you factor fully 'a4 plus 4b4?

(a2+2b2-2ab)(a2+2b2+2ab)


The eccentricity of the ellipse if minor axis is equal to the distance between the foci is Answer is 1 radical 2 how?

The standard equation for an ellipse centered at the origin is [x2/a2] + [y2/b2] = 1 We also have the relationship, b2 = a2 - c2 where c is the distance of the foci from the centre and a & b are the half lengths of the major and minor axes respectively. When the length of the minor axis equals the distance between the two foci then 2b = 2c : b = c. Thus, a2 =b2 + c2 = 2b2 One of the formulae for the eccentricity of an ellipse is, e = √[(a2 - b2)/a2] Thus, e = √[(2b2 - b2) / 2b2] = √½ = 1/√2.


Reduce -8 plus b2 by 5 plus b2?

(-8 + b2) - (5 + b2) = -8 + b2 - 5 - b2 = -13


Simplify 4a 2b a plus b?

= 4a2 + 2ab 2b2


Factor 2b2 plus 10b plus 12?

The Answer to this question is 9 (2B+4) and (1B+3)


Factor 3ab plus 3ac plus 2b2 plus 2bc?

3a(b+c)+2b(b+c)


If a2 plus b2 plus c2 - ab - bc - ca equals 0 then prove a equals b equals c?

a2 + b2 + c2 - ab - bc - ca = 0 => 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0 Rearranging, a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2 = 0 => (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0 or (a - b)2 + (b - c)2 + (c - a)2 = 0 so a - b = 0, b - c = 0 and c - a = 0 (since each square is >=0) that is, a = b = c