Divide. (2b^2 + 2b - 9)/(b-3)
2b2 + 2b - 9
———————————— =2b+8
b - 3
Is this the correct answer? Please help show me the right way if this is wrong. Thank you!!
********2 b + 8
(b-3) | 2 b^2 + 2 b - 9
********2 b^2 - 6 b
------------------------
************** + 8 b - 9
************** + 8 b - 24
-------------------------
*****************R = + 15
In other words I suspect you have a typo or you have a remainder of 15
To divide the polynomial (2b^2 + 2b - 9) by (b - 3), we can use long division. Here's how you can do it:
1. First, set up the long division problem with the dividend (2b^2 + 2b - 9) as the numerator and the divisor (b - 3) as the denominator.
___________________
b - 3 | 2b^2 + 2b - 9
2. Divide the first term of the numerator (2b^2) by the first term of the denominator (b). The result is 2b, which will be placed on top of the division symbol.
2b
_____________
b - 3 | 2b^2 + 2b - 9
3. Multiply the divisor (b - 3) by the quotient (2b). The result is 2b^2 - 6b. Write this underneath the numerator and subtract it from the original numerator.
2b
_____________
b - 3 | 2b^2 + 2b - 9
- (2b^2 - 6b)
_______________
8b - 9
4. Bring down the next term from the numerator, which is 8b. Now we have 8b - 9 as the new numerator.
2b + 8
______________
b - 3 | 2b^2 + 2b - 9
- (2b^2 - 6b)
_______________
8b - 9
5. Repeat the process by dividing the first term of the new numerator (8b) by the first term of the denominator (b). The result is 8, which will be placed on top of the division symbol, next to the previous quotient.
2b + 8
______________
b - 3 | 2b^2 + 2b - 9
- (2b^2 - 6b)
_______________
8b - 9
- (8b - 24)
_______________
15
6. Since there are no more terms to bring down, the final quotient is 2b + 8, with a remainder of 15. So the correct answer to the division problem is:
(2b^2 + 2b - 9)/(b - 3) = 2b + 8 + 15/(b - 3)
Therefore, the answer you provided, 2b + 8, is incomplete. You need to include the remainder term as well, which is 15/(b - 3).