algebra two math question?

Question

I am allowed to use any source I want to get the answer except for other people. So if you could please just tell me what I should look under in a math book, something specific, so that I can figure out exactly how I need to solve this problem that would be great. Thanks. Here is the problem. " Find all of the values of u that satisfy the equation (2u+2)(u^2+2u-24)". thanks ever so.

your right, it equals 0.

Answer 1

factor the polynomial: (u^2+2u-24) = (u + 6)(u - 2)


so now you have: (2u+2)(u+6)(u-2) = 0

you solve this, the same way you solve any other factored polynomial. set each factor to zero individually and solve:

2u+2 = 0 -> u = -1
u+6 = 0 -> u = -6
u-2 = 0 -> u = 2


as for where to look in your book. i have no idea...

Answer 2

First, since the equation = 0 then either (2u+2) = 0 or (u^2-2u-24) = 0. I would try to factor each.

2u+2=0
2(u+2) = 0
u=-2

u^2+2u-24
(u+6)(u-4) = 0
u=-6 or 4

Answer 3

im assuming the equation equals 0. you would need to look under factoring in the book. factor the second term, u^2+2u-24, and then set each ( ) equal to 0.

Answer 4

Assume question should read:-
(2u + 2).(u² + 2u - 24) = 0
2.(u + 1).(u + 6).(u - 4) = 0
u = - 1, u = - 6, u = 4
In a maths. book would come under heading of "Quadratic Equations".

Answer 5

7p + 26 = 2 - 5p Take the p's to the left factor 12p + 26 = 2 Take the numbers to the appropriate factor 12p = -24 p = -2 5(2x + 3) = 2(40 - x) boost the brackets 10x + 15 = eighty - 2x Do an analogous from first question 12x = sixty 5 x = sixty 5/12

Answer 6

I think you're missing part of the equation. There needs to be an equals sign in there somewhere.

Answer 7

try under factorising quadratic equations.

Answer 8

could i explain how to work it out??