Algebra

[iNerd]

If 2x^2 + 4x + 5 = a(x+b)^2 + c find a,b,c

I expanded it out and all, tried to equate the co-efficients but failed....:S

[m@tty]

You're going about it the hard way.

Complete the square on the left side. Done.

[iNerd]

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

[Water]

Expand a(x+b)^3 , then match them together. No easy way out.

[m@tty]

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:

[iNerd]

Thanks m@tty! Correction ...it should be -3x^2 not -3x in your third line.

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.

[m@tty]

Prove







equate coeffs:

a = p ; b = q-2p ; c = p-2q ; d = q

Then sub for a and d

QED

[monkeywantsabanana]

Q: Prove that, if ax^3 + bx^2 + cx + d = (x-1)^2(px+q) then b = d - 2a and c = a - 2d

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.

[m@tty]

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

[iNerd]

Split into partial fractions.

             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...

[brightsky]

Simplify it: (x^2 + 2x - 13)/[[2(x+3)(x^2 + 1)]
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.

EDIT: woops thanks m@tty

[m@tty]

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.

[m@tty]

Oh, and wiki it up -- http://en.wikipedia.org/wiki/Partial_fraction

Always helps.

[gossamer]

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

[m@tty]

It isn't.. hence why I added

 "And the matrices they have are just another way to solve the simultaneous equations. Don't be confused."