someone solve this maths for me
- Thread starter knip
- Start date
Get ya.
Wouldn't have a clue how to write out by hand though. I just assumed that there was one single x number and the roots made it expand ie. 3x(cubed) = 9x.
In all fairness, I doubt I could work them out by hand. It's been a good few years since I was expected to solve a cubic by hand.
Only way to do it by hand is to spot a root and then solve the remaining quadratic. That can be done normally by staring at it or worse case:
x = (-b +/- sqrt(b^2 - 4*a*c)) / 2a
Here the root to spot is x = 1, which is easy to check, 3 + 11 - 6 - 8 = 0, etc.
We always use to do them by spotting a root, but it is possible to do them another way...well the link I posted seems give another way.
I don't think ive ever been asked to solve one without an easy root to spot. These days every equation I do on my degree always has an imaginary part to it, making the task much more difficult, im glad we don't get many cubics.
For the first one you can do f(1)=0. therefore you can divide by (x-1)
(ax^2+bx+c) * (x-1) = 3x^3 +11x^2 -6x-8
(3x^3 +11x^2 -6x-8) / (x-1) ----> long division I can't be bothered to write out, she should know it though, or use the "juggling" method.
= (3x^2 + 14x + 8)
so,
(x-1)(3x^2 + 14x + 8) = 3x^3 + 11x^2 -6x - 8
You then do the quadratic formula to get the remaining values of x (-b +/- root(b^2-4ac) / 2a ) tell her to look that up.
So you get something like
(x-1)(x+b)(x+c)
X will be 1, -b, -c.
Do the same for the second and job's a good'un
(ax^2+bx+c) * (x-1) = 3x^3 +11x^2 -6x-8
(3x^3 +11x^2 -6x-8) / (x-1) ----> long division I can't be bothered to write out, she should know it though, or use the "juggling" method.
= (3x^2 + 14x + 8)
so,
(x-1)(3x^2 + 14x + 8) = 3x^3 + 11x^2 -6x - 8
You then do the quadratic formula to get the remaining values of x (-b +/- root(b^2-4ac) / 2a ) tell her to look that up.
So you get something like
(x-1)(x+b)(x+c)
X will be 1, -b, -c.
Do the same for the second and job's a good'un
She said to say this (have copied it from MSN)
its the long divison bit im strugling with i have worked out that x-1=o to divide into the some and once i have the answer to that i can do the equation to get the other 2 answers its the long division bit im stuck on!
its the long divison bit im strugling with i have worked out that x-1=o to divide into the some and once i have the answer to that i can do the equation to get the other 2 answers its the long division bit im stuck on!
ah k.
I'll edit this once I've typed it out.
Long division
I'll also type up an easier way to do it.
btw.
http://www.mathsnetalevel.com/module.php?ref=C2
login : feb
pass : feb
Do the 2 long divisions, then do the two "avoiding long divisions"
The juggling way goes like this:
3x^3 +11x^2 -6x-8 =0
(x-1)(ax^2 + bx + c) = above function. = 0
x^3 : 3 = a
x^2 : 11= b-a -> b = 14
x: -6 = -b + c -> c= 8
so,
(x-1)(3x^2 + 14x + 8)
What you do is equate the coefficients of the function and the divided function to find what the as, bs and cs are. Need any help ask more.
This is the easiest way, and is what geuben has done on his scanned version, which might be clearer - typing is pretty confusing andhard with maths :/
I'll edit this once I've typed it out.
Long division
I'll also type up an easier way to do it.
btw.
http://www.mathsnetalevel.com/module.php?ref=C2
login : feb
pass : feb
Do the 2 long divisions, then do the two "avoiding long divisions"
The juggling way goes like this:
3x^3 +11x^2 -6x-8 =0
(x-1)(ax^2 + bx + c) = above function. = 0
x^3 : 3 = a
x^2 : 11= b-a -> b = 14
x: -6 = -b + c -> c= 8
so,
(x-1)(3x^2 + 14x + 8)
What you do is equate the coefficients of the function and the divided function to find what the as, bs and cs are. Need any help ask more.
This is the easiest way, and is what geuben has done on his scanned version, which might be clearer - typing is pretty confusing andhard with maths :/
Last edited:
Soldato
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Get her to have a look at one of these:
http://www.sosmath.com/algebra/factor/fac01/fac01.html
http://en.wikipedia.org/wiki/Polynomial_long_division
They're fairly good at explaining long division
http://www.sosmath.com/algebra/factor/fac01/fac01.html
http://en.wikipedia.org/wiki/Polynomial_long_division
They're fairly good at explaining long division
Ok so im trying to help out a mate whos had a really stressful week, can someone solve these two maths puzzles for me
3x 3 + 11x 2 -6x -8 =0
2x 3 +13x 2 -24x +9=0
The little 3 and 2 are squared and cubed.
Please help!
Fanks
first one :- x= 1 x= -2/3 x = -4
second one :- x=1 x= 0.5584 x=-8.058
ah k.
I'll edit this once I've typed it out.
Long division
I'll also type up an easier way to do it.
btw.
http://www.mathsnetalevel.com/module.php?ref=C2
login : feb
pass : feb
Do the 2 long divisions, then do the two "avoiding long divisions"
The juggling way goes like this:
3x^3 +11x^2 -6x-8 =0
(x-1)(ax^2 + bx + c) = above function. = 0
x^3 : 3 = a
x^2 : 11= b-a -> b = 14
x: -6 = -b + c -> c= 8
so,
(x-1)(3x^2 + 14x + 8)
What you do is equate the coefficients of the function and the divided function to find what the as, bs and cs are. Need any help ask more.
This is the easiest way, and is what geuben has done on his scanned version, which might be clearer - typing is pretty confusing andhard with maths :/
She said to say
You are an absolute angel you dont know how happy you have just made me xxxx
(I fink that means fanks!)
Awww all the effort (ok it wasn't that much) and I didn't get recognition
...and i posted first.Awww all the effort (ok it wasn't that much) and I didn't get recognition
Ok she loves both of you then!
ah k.
I'll edit this once I've typed it out.
Long division
you missed out the ^2 on the first line under numerator.
