- #1
randoreds
- 24
- 0
Ok all save y'all a little reading.
Worked out the problem. Got
X^2+2x-1=A(2x-1)(x+2)+bx(x+2)+cx(2x-1)
Ok then you write it standard form for a polynomial. Then use can use there coefficients to write new equations at you get
2a+b+c=1
3a+2b-c=2 and finally
-2a=1
Now you solve for a,b,c and this is where I'm confused.
To solve for a is easy, it is just 1/2. But the book doesn't show how to solve for b and c.
B =1/5 and C =-1/10
I'm totally confused how they got those numbers so if you could explain it. That would be so helpful!
Worked out the problem. Got
X^2+2x-1=A(2x-1)(x+2)+bx(x+2)+cx(2x-1)
Ok then you write it standard form for a polynomial. Then use can use there coefficients to write new equations at you get
2a+b+c=1
3a+2b-c=2 and finally
-2a=1
Now you solve for a,b,c and this is where I'm confused.
To solve for a is easy, it is just 1/2. But the book doesn't show how to solve for b and c.
B =1/5 and C =-1/10
I'm totally confused how they got those numbers so if you could explain it. That would be so helpful!