Please help! Show me how to factor these binomials! m^4-256 and z^3+125, thank you! Descriptions are appreciated :)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Please help! Show me how to factor these binomials! m^4-256 and z^3+125, thank you! Descriptions are appreciated :)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

m^4-256 this is a perfect square
because 2*2=4, right?
the 1 you want to concentrate on is 256

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

ok
try and get it into the form \(a^2-b^2\) Use the hints that mth3v4 has given you.
cool pic asna :D
thx :) I'm actually an alien from another planet :D
:)
Starshine7 - can you express 256 as (something)^2 ?
16x16?
:)
good, now can you express \(m^4\) as (something)^2 ?
so can you put that in in a factored form
m*m*m*m?
Yes, so \(m^4=(m^2)^2\) Now use those two to get the expression into the form \(a^2-b^2\) so that you can then factor it like this:\[a^2-b^2=(a+b)(a-b)\]
why is it written like (m+number)(trinomial)? final answer i mean. If it helps, I factor by grouping if i can incorporate that somehow
just do this one step at a time and al will be revealed :)
ok
lol oops
nothing happened
:)
Starshine7 - are you stuck?
Sadly yes :(
look at the first answer you gave me
how did you come up with that ?
ok, we saw above that:\[256=16^2\]\[m^4=(m^2)^2\]so now we can rewrite your original equation as:\[m^4-256=(m^2)^2-16^2\]now mak use of:\[a^2-b^2=(a+b)(a-b)\]to simplify this further.
after that if can be factored again as an option as you can see
(m^2)^2 confuses me
(i think i am confusing starshine)
ok, lets use a substitution to make it clearer...
let \(n=m^2\), then we have:\[m^4-256=(m^2)^2-16^2=n^2-16^2\]
why n? o.O
we could use any other symbol - it doesn't really matter
the point is it is now in the form \(a^2-b^2\)
so you should now be able to factorise it.
okies properties of exponents state when you multiply exponent and exponent you add the exponent values together (a^x)(a^b)=a^x+b
exponent properties/laws of exponents whatever
Starshine7 - are you familiar with the exponent laws mth3v4 has shown? i.e.:\[(x^a)^b=x^{ab}\]
If it helps I dont understand when a and b are examples, I was taught using only the numbers, so thats why im really confused :(
those are just any number
I know >_> Im weird like that
so you are comfortable with something like this:\[18^2-16^2=(18+16)(18-16)\]but when replaced with symbols like this:\[a^2-b^2=(a+b)(a-b)\]you get lost?
Yes, exactly :)
ok - the 'a' and 'b' are just placeholders for any number
I just have attention issues so the numbers help me understand it better, i drift easily.
what this says is that the formula:\[a^2-b^2=(a+b)(a-b)\]is true for ANY value of 'a' and 'b'
so in your example above, we ended up with:\[m^4-256=(m^2)^2-16^2=n^2-16^2\]which means we can think of it as a=n and b=16 to get:\[n^2-16^2=(n+16)(n-16)\]we can then substitute back \(n=m^2\) to get:\[m^4-256=(n+16)(n-16)=(m^2+16)(m^2-16)\]
do you understand those steps?
I think so
(a ^x ) ( a ^b )= a^x+b ^ ^ ^ ^ ^ ^ ^ | | | | / | | | L _____|___ L _ |___L_ _ L _ exponent number both x and b can be any number | | | your number ______ |
does my funny drawing help XD
ok, so now you should notice that we have \((m^2-16)\) as one of the factors. since \(16=4^2\), we can write this as:\[(m^2-16)=(m^2-4^2)=(m+4)(m-4)\]using the same formula. we therefore end up with:\[m^4-256=(m^2+16)(m+4)(m-4)\]
So you just keep factoring to the lowest numbers that are perfect squares?
yes
mth3v4 - you have a lot of patience to draw all that :)
lol
:)
it still comes out a bit messy
(a^x) (a^y) = a ^ x+y lemme try this you have this thing here right starshine?
Starshine7 - for your second expression, you need to use the following factorisation:\[a^3+b^3=(a+b)(a^2-ab+b^2)\]so you need to get your expression:\[z^3+125\]into the form \(a^3+b^3\)
okies finish what you have first
I guess
I have to go now guys - Starshine I am sure mth3v4 will help you with the rest. Good luck all...
Aww..well bye Asna :) thanks anyways
kk bb
yw
z^3+125 is your next 1
Yes
it is not going to be squared but cubed
have you heard of diff of cubes or sum of cubes formulas
no
so what ansa wrote there before leaving was formula for diff of cubes
because you just see just a 2 termed expression here
so how would you make x^3 and change it to x^2
better yet lemme ask the question another way how do you get x^3 from x^2
I'm not sure. I think they only have 2 exs in common
okies lets try this ( a^x ) ( a^y ) = a ^ x +y the variable "a" can be any number 1, 2, 3, whatever, happy face , i dont care what you want it to be ok so far
this just goes fot the variable "a" we good?
yes
then next this for the variables "x, y" these give a numeric value (these have to be a number) giving a value ( a^x ) ( a^y) = a ^ x + y
so in the end the numeric value of your "a" exponent value "x" "a" exponent value "y" they add up result for the a ^ x + y is this more understandable?
Yes
so can we try that again how to get x^3 from x^2
(x)(x+x)?
i just want it from x^2
I'm sorry I have no idea what to do :(
since it is already squared right (x^2)(x)=x how many more "x" do we need for x ^3
x^2
you have x^2 already
but x^3 is made up of x^3 theres not anything left to do
remember (x^2)(x)=x? ^ | value of exponent you are adding to the result
lol if there is nothing then say nothing
be more sure of you self now just give the reason
nm that its probably more difficult to explain
but do you understand how it works
thats all i want to know
How about we start over with a new problem, m^3+216n^3 Show me in steps how to solve this and ill try another on my own, this might work
okeedokee
lol this is differnt now because it has 2 variables in it
okie now we can use something called difference of squares a^3+b^3= (a+b) (a^2-ab+b^2)
ok
first of all you must clearly understand the laws of exponents (or at least the, 1 i showed you, you will have to use it for these problems )
we can go over it again i dont mind
sure :)
okie i will put a sample problem (x^3) (x^3 )=x ^6 the exponents values of "x " add up together (d ^100) ( d ^ 90) = d ^ 190 exponents values of "d" add up together
does this make sense :)
yes
"for now" <--------------------note i can go over with the rest of the laws but i hhave limited time this is what you will need difference of squares a^3+b^3= (a+b) (a^2-ab+b^2
ok
m^3+216n^3 know what you are working with m^3 + 216n^3 ^ ^ first term | second term
a^3+b^3= ( a+b ) (a^2- ab+ b^2) ^ ^ you only have "items" that you are going to be multiplying using FOIL with
a^3+b^3= (a+b) (a^2-ab+b^2) | ^ ^ | this part wants the numbers squared to | be multiplied with L--------------------|
how do we get a squared variable to be cubed? :)
^3?
m^2 to m^3
add another m?
type how you would do that plz :)
its correct but how would it look
mathematically
give you a hint its not x+x
that is called adding like terms
it becomes 2x
remember this model (a ^ x) ( a ^y ) = a ^x+y
x is going to be your a variable
the values of your exponent to the x are?
i gave you already x^2 you have x^2 how much you need to get to x^3
(x ^ 2) ( x ^y ) = x ^2+y what will you need to get 3
as the exponent result
x
correct :)
so the value of that 1 that you are missing is just x
^^
it is because x*x*x = x^3 but i think you know that
do you know how to simplify radicals
I do
im sorry i have chat lag :(
cube root of 216 is 6 but you want 6^2 and n^2 so you can do the FOIL
sorries i gtg now :(

Not the answer you are looking for?

Search for more explanations.

Ask your own question