BEC'S methods questions

[Mao]

aiisha u do know you can edit your own posts

(Image removed from quote.)

lol


on the side note, i think factorising (when possible) is a lot easier than the general formula, as for this case you're dealing with whole numbers, i'd rather only do a couple mental arithmetics than surds (somewhat)

[Collin Li]

Oh yeah, factorising is very easy too. Going from shouldn't even need working out. I was comparing it to completion of the square (which is what she tried).

[bec]

ahh thanks both of you, pretty obvious i guess!


last one:

Solve for x:

x⁴ − x³ + x − 1 = 0
I did it in my head to get x=1, -1
What would the "method" be if you had to show working? Could you just write something like...

LHS: Try x=1
(1)⁴ -1³ +(1) = 1 - 1 + 1
                  = RHS, therefore x=1 is a solution

Let x=-1
(-1)⁴ -(-1)³ +(-1) = 1 + 1 - 1
                        = RHS, therefore x=-1 is a solution

[Collin Li]

No, that wouldn't be sufficient because how do you know if there are no other solutions (there can be up to 4 since it is a degree 4 polynomial)?

You should be factorising instead. You would write something like:






Then, you know that therefore is a factor, then you can do long or "synthetic" division to get a quadratic, and possibly factorise that too.



Now we can confirm with the discriminant that the quadratic factor, has no real solutions:




So now you can confidently say that are all of the real solutions to

[bec]

gahh ok yes yes i understand, i think my factorisation skills have been slowly dying for a while now.
you've started a resurrection

[droodles]

bec are u doing polynomials in the first term or something, i thought most schools started with trig

[AppleXY]

Exactly.

Since you already found 2 factors, (x+1)(x-1) (derived by your solutions x=1,-1), you can use long ("synthetic") division to obtain a quadratic factor by expanding it out.

But as x^2 - x + 1 doesn't have any real solutions you must leave it like that.

Or you could use equality of polynomials.

(x+1)(ax^2+bx+c) = f(x)

Equate. gotta run now

[Collin Li]

Also, here's a much easier way to do it by grouping:



Using :

Or you could use equality of polynomials.

(x+1)(ax^2+bx+c) = f(x)

Equate. gotta run now

What he means is this:

We know that , and that is a factor.

Therefore we know that can be written as

Expanding the LHS, we get:

[/0]

I think you meant , not .

Sorry had DOPS in my head

[bec]

what am i doing wrong here? it's a cas question btw:

Five numbers add to zero. The first is equal to the sum of the second and the fourth. The third is equal to the sum of the fourth and the fifth. The sum of the first two numbers is 2 more than the fifth number. The fifth number is three times the sum of the third and fourth numbers. Find the five numbers.

Let a=1st number, b=2nd etc etc

a+b+c+d+e=0
a=b+d
c=d+e
a+b=2+e
e=3(c+d)

Using rref on my calc, i keyed in this matrix:

1  1  1  1  1  0
1 -1  0 -1 0  0
0  0  1 -1 -1 0
1  1  0  0 -1  0
0  0  3  3  1  0

and got a single solution matrix which told me that:
a=4/5,  b=0,  c = -2/5,  d=4/5,  e=-6/5

only, the solution in the book says the numbers are:
5/7,  3/7,  -4/7,  2/7,  -6/7

Anyone have any idea what i'm doing wrong?
thanks

[iamdan08]

The matrix set up should be:
                       a b  c  d  e

                       1  1  1  1  1  0
                      -1  1  0  1  0  0
                       0  0 -1  1  1  0     -> (sto)        then rref                then you will get the answers in your book
                       1  1  0  0 -1  2
                       0  0 -3 -3  1  0

All you have done wrong is set up the matrix incorrectly.

You need to have the coefficents for a b c d e in the correct order in the matrix.

If there is no e value for example, then you enter 0 into the matrix.


Not sure if that makes much sense, but yeah... lol

[iamdan08]

If for example you had:

 a=b+d

 transpose to -1a+1b+0c+1d+0e=0

you would enter
   a  b  c   d  e
 -1   1  0   1  0    0   


Hope that helps!!!

[Mao]

a+b+c+d+e=0
a=b+d
c=d+e
a+b=2+e
e=3(c+d)

makes the matrix

which is a little different to your matrix =\

which yields the right answer check your matrix, you'll find your 4th row should end with 2, and your 5th row should have -3s instead of 3s

[bec]

thanks iamdan. that helps, i realised i just accidently had 3's in my bottom row instead of -3s.
the other values you got different to me are just because we transposed things differently (but we're both right)
eg.

If for example you had:

 a=b+d

 transpose to -1a+1b+0c+1d+0e=0

With this one i just transposed it to a-b-d=0 (and now that i fixed the 3 thing, got the right answer)

edit: mao you're right, i had the 2 in my matrix but just didn't type it up right here

[bec]

also i've got yet another question... it's probably really simple but how do i do this?

A quartic function has equation y = ax⁴ + bx³ + c² + dx + e.  Its graph cuts the x-axis at (-1,0) and (2,0). One of these intercepts is a stationary point of inflection. If the graph passes through (1,16), find a,b,c,d and e.

thanks!