calculusxy
  • calculusxy
factoring quadratic expressions
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
calculusxy
  • calculusxy
\[18n^2 - 8\]
calculusxy
  • calculusxy
@phi
Nnesha
  • Nnesha
take out the common factor what is GCF(greatest common factor ) ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

calculusxy
  • calculusxy
2?
Nnesha
  • Nnesha
right take out 2 from 18n^2 -8 or in other words divide both terms by common factor \[2(\frac{ 18n^2 }{ 2 }-\frac{ 8 }{ 2 })\]
calculusxy
  • calculusxy
2(9n^2 - 4)
phi
  • phi
you should always be on the look out for "difference of squares"
calculusxy
  • calculusxy
@phi I don't understand...
Nnesha
  • Nnesha
9 and 4 are perfect square roots and the negative between both terms so you can apply the difference of squares \[\huge\rm a^2-b^2 =(a-b)(a+b)\]
calculusxy
  • calculusxy
so how would i implement it on 9^2 - 4?
Nnesha
  • Nnesha
this is how i do it take square roots of both terms write in two parentheses (sqrt of 1st term `+` sqrt of 2nd term) (sqrt of 1st term `-` sqrt of 2nd term ) lolol
calculusxy
  • calculusxy
okay.. so would it be (3+4)(3-4)?
Nnesha
  • Nnesha
what about n it's 9n^2 - 4
calculusxy
  • calculusxy
(3n+4)(3n-4)
Nnesha
  • Nnesha
looks good don't forget the common factor 2(3n+4)(3n-4)
Nnesha
  • Nnesha
ohh wait no
Nnesha
  • Nnesha
|dw:1444163520807:dw| take square root of both termz
Zarkon
  • Zarkon
\[9n^2 - 4 =(3n)^2-2^2\]
calculusxy
  • calculusxy
sorry but this is not clear to me at all
calculusxy
  • calculusxy
i need more help on this
Nnesha
  • Nnesha
alright let's start when the coefficient of both terms are perfect squares and the exponent of the variable is even and the negative sign between the terms then you can use difference of squares hint
anonymous
  • anonymous
im not that good at math. I would try to help you but I would end up failing.
anonymous
  • anonymous
no thats not true you wont fail.
anonymous
  • anonymous
and if you do like there is a saying First comes frailer second comes success.
Nnesha
  • Nnesha
\[\huge\rm x^2 -4\] even power and both coefficient are perfect squares since the negative sign between them we can use the hint \[\huge\rm\color{ReD}{ a}^2 -\color{blue}{b}^2 = (a-b)(a+b)\] we have to rewrite 4 as 2 exponent so 2^2 = 4 \[\huge\rm \color{ReD}{x}^2-\color{blue}{2}^2=(\color{Red}{x}-\color{blue}{2}(\color{Red}{x}+\color{blue}{2}) \]
anonymous
  • anonymous
My god.
anonymous
  • anonymous
that wuill take a while. lol.
Nnesha
  • Nnesha
that's what zarkon mentioned above that's same as taking square roots of both terms
anonymous
  • anonymous
ahh
Nnesha
  • Nnesha
|dw:1444164108305:dw| one with negative sign and one with positive
Nnesha
  • Nnesha
now make sense or no ?? hm
calculusxy
  • calculusxy
u kinda do.. but i m still trying to understand it
Nnesha
  • Nnesha
alright take ur time let me know what u don't understand :=) will try to explain
calculusxy
  • calculusxy
if we have 16^2 - 49, would we have (4 - 7)(4+7) or something else?
Nnesha
  • Nnesha
there should be a variable
calculusxy
  • calculusxy
where?
Nnesha
  • Nnesha
it should be 16x^2 -49
Nnesha
  • Nnesha
remember if it's 16^2-49 both are `like` terms so we can just combine them
calculusxy
  • calculusxy
oh okay... so would we have that as (4n + 7)(4n - 7) ?
Nnesha
  • Nnesha
looks good
calculusxy
  • calculusxy
so for my original problem the answer would be:
calculusxy
  • calculusxy
\[2(9n^2 - 4)\] \[9n^2 - 4 \implies (3n + 2)(3n-2)\] \[\large Answer\] \[2(3n+2)(3n-2)\]
Nnesha
  • Nnesha
\(\huge\color{Green}{\checkmark}\)
Nnesha
  • Nnesha
that's right!
calculusxy
  • calculusxy
Thank you so much! I just want to review the steps...
Nnesha
  • Nnesha
just do some example u will get the concept :=)
calculusxy
  • calculusxy
1. Find the GCF of the terms 2. Put the GCF outside of the parenthesis. 3. Divide the terms of the question by the GCF. 4. (if there are perfect squares, then we use a^2 - b^2 = (a + b)(a - b)
calculusxy
  • calculusxy
5. and then put the GCF outside
Nnesha
  • Nnesha
right for perfect squares: sign should be negative between terms both numbers should be perfect squares even power!
calculusxy
  • calculusxy
i have a small question... if we have -x^2, can we turn the x to be a positive ?
calculusxy
  • calculusxy
...since the exponent is a positive, i thought that that's possible
calculusxy
  • calculusxy
@Nnesha
Nnesha
  • Nnesha
it depends on 2nd term let's say we have \[\huge\rm -x^2 - 9\] we can factor out the negative one \[\large\rm -1(x^2+9)\] now there is a positive sign we can't factor it out
Nnesha
  • Nnesha
and if we have \[\huge\rm -x^2+9\] take out the negative one a should be positive
calculusxy
  • calculusxy
-x^2 + y^3... i think this was a question on a test that i took
calculusxy
  • calculusxy
like we had to divide with exponents and that was in the denominator... wait. let me write it out \[\frac{ 1 }{ -x^{-2}y^3 }\]
calculusxy
  • calculusxy
i moved -x^{-2} up to the numerator. \[\frac{ -x^2 }{ y^3 } \implies \frac{ x^2 }{ y^3}\]
calculusxy
  • calculusxy
would that be valid?
Nnesha
  • Nnesha
ohhh no wait why did you flip the fraction ?
Nnesha
  • Nnesha
IF IF `exponent ` is negative then you should flip the fraction
calculusxy
  • calculusxy
i only moved -x^{-2} since the exponent was negative
Nnesha
  • Nnesha
wait a sec its lagging let me refresh the page
Nnesha
  • Nnesha
\[\huge\rm \frac{ 1 }{ -x^2y^3}\] is this ur question ?
calculusxy
  • calculusxy
no\[\huge \frac{ 1 }{ -x^{-2}y^3 }\]
Nnesha
  • Nnesha
ohh okay make sense
Nnesha
  • Nnesha
right \[\huge\rm \frac{ 1 }{ -x^{-2}y^3 } =- \frac{ x^2 }{ y^3 }\] sign would be at the top never leave the negative sign in the denominator negative times positive = negative
Nnesha
  • Nnesha
you forgot the negative sign variables are correct :=)
calculusxy
  • calculusxy
ok thank you!
Nnesha
  • Nnesha
np :=) good luck!

Looking for something else?

Not the answer you are looking for? Search for more explanations.