SOLUTION: pg. 374 no. 70. -3s^2-10s+8 I can't get my signs to be correct in order for the 10 to be negative. What am I doing wrong?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: pg. 374 no. 70. -3s^2-10s+8 I can't get my signs to be correct in order for the 10 to be negative. What am I doing wrong?      Log On


   



Question 54672This question is from textbook Beginning Algebra
: pg. 374 no. 70.
-3s^2-10s+8
I can't get my signs to be correct in order for the 10 to be negative. What am I doing wrong?
This question is from textbook Beginning Algebra

Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
-3s2-10s+8 
I can't get my signs to be correct in order for the 
10 to be negative. What am I doing wrong?

-3s2 - 10z + 8

Whenever the first term is negative, always begin by
factoring out -1, which will change all the signs
inside the parentheses.

-1(3s2 + 10s - 8)

1. Multiply 3 by 8, getting 24

2. Think of two positive integers whose product is 24
and whose difference (since the last sign is negative)
is 10.  These are 12 and 2.

3. Attach signs to 12s and 2s so they will combine to
give the middle term +10s.  So we attach + to the 12s 
and - to the 2s

4. Rewrite the middle term +10s as +12s - 2s.  Change 
the parentheses to brackets, since they will soon
contain parentheses themselves.

-1[3s2 + 12s - 2s - 8]

5. Factor the first two terms in the brackets by taking 
out 3s.  Factor the last two terms in the brackets by 
taking out -2

-1[3s(s + 4) - 2(s + 4]]

6. Factor (s + 4) out of the two terms in the brackets:

-1[(s + 4)(3s - 2)]

7. Erase the brackets

-1(s + 4)(3s - 2)
 
8. Dispense with the 1 but leave the negative sign:

-(s + 4)(3s - 2) 
 
Edwin