Can (y+3)^3 + 8 Be Fully Factorized?

[alpha01]

$$(y + 3)^3 + 8$$


my attempt has led me to

(y+3)(y+3)(y+3+8)

but i doubt this is correct

 

[tiny-tim]

Hi alpha01!

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 $$z^3 + 8$$.

Can you do that?

(Hint: when is it zero?)

 

[alpha01]

expanding and simplify the (y+3)^3:

(y^2 + 6y + 9)(y + 3) so is the answer:

(y^2 + 6y + 9) (y + 3 + 8)?

 

[alpha01]

Please post the solution and i will apply it to other examples

 

[tiny-tim]

expanding and simplify the (y+3)^3:

(y^2 + 6y + 9)(y + 3)

No no no … that is expanding but it's not simplifying!

"Expanding" means "longer", and "simplifying" means "shorter"!

$$z^3 + 8$$ is shorter and simpler than $$(y + 3)^3 + 8$$.

So: can you factorise $$z^3 + 8$$?

 

[alpha01]

(z+2)(z+2)(z+2)

so the answer to the original question is:

(y+3+2)(y+3+2)(y+3+2)

(y+5)(y+5)(y+5)?

 

[tiny-tim]

(z+2)(z+2)(z+2)

No.

Hint: can you see a number for which $$z^3\,+\,8\,=\,0$$?

(In other words, a root of $$z^3\,+\,8\,=\,0$$)

Put "z - " in front of it, and that gives you a factor of $$z^3\,+\,8\,=\,0$$!

(if you still can't get it, I'll give you the answer)

 

[alpha01]

no..

 

[Feldoh]

Solve for z:
$$z^3\,+\,8\,=\,0$$

You need to think of this as

$$z^3+x^3 = 0$$

Where x^3 = 8

 

[alpha01]

z = -6

 

[Feldoh]

How are you getting z = -6?

$$z^3\,+\,8\,=\,0$$

$$z^3\,=\,-8$$

$$z\,=\,(-8)^{1/3}$$

 

[kbaumen]

To solve this one, you should know this formula: a^3 + b^3 = (a + b)(a - ab + b). I hope that helps.

 

[alpha01]

To solve this one, you should know this formula: a^3 + b^3 = (a + b)(a - ab + b). I hope that helps.

thank you.. that's all i needed

 

[alpha01]

How are you getting z = -6?

$$z^3\,+\,8\,=\,0$$

$$z^3\,=\,-8$$

$$z\,=\,(-8)^{1/3}$$

i meant to write -2

i know how to solve this now anyway

 

[alpha01]

So, putting (y+3)^3 +8 in the form (a+b)(a^2 -ab+b^2):

$$((y+3)+2) ((y+3)^2 -2(y+3)+4)$$


but i don't think this is completely factorized?

 

[Gib Z]

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.

 

[tiny-tim]

So, putting (y+3)^3 +8 in the form (a+b)(a^2 -ab+b^2):

$$((y+3)+2) ((y+3)^2 -2(y+3)+4)$$

but i don't think this is completely factorized?

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?

 

[alpha01]

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.

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.

 

[alpha01]

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.

ok here's 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.

 

[HallsofIvy]

Your original post was:

$$(y + 3)^3 + 8$$


my attempt has led me to

(y+3)(y+3)(y+3+8)

but i doubt this is correct

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!

 

[Schrodinger's Dog]

So, putting (y+3)^3 +8 in the form (a+b)(a^2 -ab+b^2):

$$((y+3)+2) ((y+3)^2 -2(y+3)+4)$$


but i don't think this is completely factorized?

Simplify this and you will have your answer.