[tex](y + 3)^3 + 8[/tex]
my attempt has led me to
but i doubt this is correct
If the answer isn't obvious, then simplify the question as much as possible, then try to solve it, then "un-simplify" it back.
Put z = y + 3.
Then the question is: factorise [tex]z^3 + 8[/tex].
Can you do that?
(Hint: when is it zero?)
expanding and simplify the (y+3)^3:
(y^2 + 6y + 9)(y + 3)
so is the answer:
(y^2 + 6y + 9) (y + 3 + 8)?
Please post the solution and i will apply it to other examples
No no no … that is expanding but it's not simplifying!
"Expanding" means "longer", and "simplifying" means "shorter"!
[tex]z^3 + 8[/tex] is shorter and simpler than [tex](y + 3)^3 + 8[/tex].
So: can you factorise [tex]z^3 + 8[/tex]?
so the answer to the original question is:
Hint: can you see a number for which [tex]z^3\,+\,8\,=\,0[/tex]?
(In other words, a root of [tex]z^3\,+\,8\,=\,0[/tex])
Put "z - " in front of it, and that gives you a factor of [tex]z^3\,+\,8\,=\,0[/tex]!
(if you still can't get it, I'll give you the answer)
Solve for z:
You need to think of this as
[tex]z^3+x^3 = 0[/tex]
Where x^3 = 8
z = -6
How are you getting z = -6?
To solve this one, you should know this formula: a^3 + b^3 = (a + b)(a - ab + b). I hope that helps.
thank you.. thats all i needed
i meant to write -2
i know how to solve this now anyway
So, putting (y+3)^3 +8 in the form (a+b)(a^2 -ab+b^2):
[tex]((y+3)+2) ((y+3)^2 -2(y+3)+4)[/tex]
but i dont think this is completely factorized?
I'm not trying to be rude, but from posts 1, 3 and 6 it seems to me you really need to go back and brush up on your algebra.
Well, you must then expand the two main brackets so that there's no brackets inside them!
At that point, you've finished - it can't be factorised any further.
[size=-2](Unless you know about complex numbers.)[/size]
Though, just to check, you should then multiply it out again to make sure you get back to the original!
Are you still worried about anything?
I'm not trying to be rude, but posts 1, 3, 6 are all direct attempts of factorization. My general algebraic skills are fine (19/20 of a algebra quiz (no factorization questions) supports this).
tiny-tim, thanks all solved.
ok heres post #3
(y+3)^3 expands to:
(y^2 + 6y + 9)(y + 3)
So can you please explain (with detailed solution) how this is apparently incorrect (and yes i know it can be expanded further i deliberately didn't expand it further.)
and yes i know what "simplifying" is, i didnt simplify because i hadn't yet fully expanded it and that would take it out of the ()() form, which is obviously not going to factorize it.
Your original post was:
It certainly isn't correct because you can't take the "8" inside one factor. That 8 would now be multiplied by the other "(y+ 3)" terms and it isn't in the original form.
What you should have seen immediately was that (y+ 3)3+ 8= (y+3)3+ 23. Do you know a general formula for factoring a3+ b3? If you are asked to do a problem like this, I suspect it is introduced in this section!
Separate names with a comma.