Post by: iNerd on April 10, 2011, 01:22:52 pm
I expanded it out and all, tried to equate the co-efficients but failed....:S
Post by: m@tty on April 10, 2011, 01:24:03 pm
Complete the square on the left side. Done.
Post by: iNerd on April 10, 2011, 01:32:30 pm
Post by: Water on April 10, 2011, 01:34:56 pm
Post by: m@tty on April 10, 2011, 01:38:12 pm
If 3x^3 - 9x^2 + 9x + 2 can be expressed in the form a(x+b)^3 +c then find the values of a,b,c
Well, a = 3 you can read that off the x^3 term
Then remember that (x+b)^3 = x^3+3x^2*b+3x*b^2+b^3
So you've got
You can see that b = -1
Then to finish off:
Post by: iNerd on April 10, 2011, 01:46:18 pm
Q: Prove that, if ax^3 + bx^2 + cx + d = (x-1)^2(px+q) then b = d - 2a and c = a - 2d
Q: If 3x^2 + 10x + 3 = c(x-a)(x-b) for all values of x, find the values of a,b,c.
Post by: m@tty on April 10, 2011, 01:56:01 pm
equate coeffs:
a = p ; b = q-2p ; c = p-2q ; d = q
Then sub for a and d
QED
Post by: monkeywantsabanana on April 10, 2011, 02:01:34 pm
expand the RHS
ax^3 + bx^2 + cx + d = (x^2-2x+1)(px+q)
ax^3 + bx^2 + cx + d = px^3 +qx^2-2px^2-2qx+px+q
ax^3 + bx^2 + cx + d = px^3+(q-2p)x^2(-2q+p)x+q
Equate co-efficients:
a=p
b=q-2p
c=-2q+p
d=q
therefore
b=d-2a (since d=q and a=p)
and
c=-2d+a (since d=q and a = p)
Proven.
Q: If 3x^2 + 10x + 3 = c(x-a)(x-b) for all values of x, find the values of a,b,c.
Expand the RHS
3x^2 + 10x + 3 = c(x^2-bx-ax+ab)
3x^2 + 10x + 3 = cx^2-cbx-acx+abc
3x^2 + 10x + 3 = cx^2-(cb+ac)x+abc
Equate co-efficients
c=3
-(cb+ac)=10
-c(b+a) =10
-3(b+a)=10
b+a = -10/3
b= -10/3 - a
3ab=3
ab=1
a=1/b
Simultaneous eqns.
b=-10/3 - 1/b
b= -1/3 or -3
sub back into eqns.
when b = -1/3
a=-3
when b = -3
a= -1/3
a= -3, b= -1/3, c= 3 OR a= -1/3, b= -3, c=3
Sorry - it's a bit hard to read. I don't know how to do the big math working out like others do..
and do correct me anyone if i'm wrong.
Post by: m@tty on April 10, 2011, 02:04:50 pm
I can't see any easy factors so use quadratic formula.
EDIT: Wait, 3 and 1/3 are the factors, so a = -1/3 b = -3 or vice versa
Post by: iNerd on April 11, 2011, 08:02:06 pm
x^2 + 2x - 13
_______________________________
2x^3 + 6x^2 + 2x + 6
All the partial fractions I've done so far have a numerator that's either a constant or linear...
Post by: brightsky on April 11, 2011, 08:10:17 pm
Hence the partial decomp is of the form A/(x^2 + 1) + B/2(x+3)
Just evaluate from there.
...
where A is of the form ax + b. :P
EDIT: woops thanks m@tty
Post by: m@tty on April 11, 2011, 08:11:27 pm
Then go
With partial fractions put the numerator as one degree less than the denominator.
Look on wolfram alpha for full working - MATHS MATHS MATHS
Click show steps on the partial fraction section..
And the matrices they have are just another way to solve the simultaneous equations. Don't be confused.
Post by: m@tty on April 11, 2011, 08:19:11 pm
Always helps.
Post by: gossamer on April 12, 2011, 08:26:33 pm
It equals
Then go
With partial fractions put the numerator as one degree less than the denominator.
Look on wolfram alpha for full working - MATHS MATHS MATHS
Click show steps on the partial fraction section..
And the matrices they have are just another way to solve the simultaneous equations. Don't be confused.
Just a question, we haven't done partial fractions in class yet, but the steps in your link include Gaussian elimination. I'm just a little confused because we've just done that in Uni Maths, and I was under the impression that it (gaussian elimination) wasn't taught in GMA? o_o
Post by: m@tty on April 12, 2011, 08:36:12 pm
"And the matrices they have are just another way to solve the simultaneous equations. Don't be confused."
Post by: gossamer on April 12, 2011, 08:44:07 pm