factoring quadratic expressions

- calculusxy

factoring quadratic expressions

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- calculusxy

\[18n^2 - 8\]

- calculusxy

@phi

- Nnesha

take out the common factor what is GCF(greatest common factor ) ?

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## More answers

- calculusxy

2?

- 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

2(9n^2 - 4)

- phi

you should always be on the look out for "difference of squares"

- calculusxy

@phi I don't understand...

- 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

so how would i implement it on 9^2 - 4?

- 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

okay.. so would it be
(3+4)(3-4)?

- Nnesha

what about n it's 9n^2 - 4

- calculusxy

(3n+4)(3n-4)

- Nnesha

looks good don't forget the common factor
2(3n+4)(3n-4)

- Nnesha

ohh wait no

- Nnesha

|dw:1444163520807:dw|
take square root of both termz

- Zarkon

\[9n^2 - 4 =(3n)^2-2^2\]

- calculusxy

sorry but this is not clear to me at all

- calculusxy

i need more help on this

- 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

im not that good at math. I would try to help you but I would end up failing.

- anonymous

no thats not true you wont fail.

- anonymous

and if you do like there is a saying First comes frailer second comes success.

- 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

My god.

- anonymous

that wuill take a while. lol.

- Nnesha

that's what zarkon mentioned above
that's same as taking square roots of both terms

- anonymous

ahh

- Nnesha

|dw:1444164108305:dw|
one with negative sign and one with positive

- Nnesha

now make sense or no ?? hm

- calculusxy

u kinda do.. but i m still trying to understand it

- Nnesha

alright take ur time let me know what u don't understand :=) will try to explain

- calculusxy

if we have 16^2 - 49, would we have (4 - 7)(4+7) or something else?

- Nnesha

there should be a variable

- calculusxy

where?

- Nnesha

it should be 16x^2 -49

- Nnesha

remember if it's 16^2-49 both are `like` terms so we can just combine them

- calculusxy

oh okay... so would we have that as (4n + 7)(4n - 7) ?

- Nnesha

looks good

- calculusxy

so for my original problem the answer would be:

- calculusxy

\[2(9n^2 - 4)\]
\[9n^2 - 4 \implies (3n + 2)(3n-2)\]
\[\large Answer\]
\[2(3n+2)(3n-2)\]

- Nnesha

\(\huge\color{Green}{\checkmark}\)

- Nnesha

that's right!

- calculusxy

Thank you so much!
I just want to review the steps...

- Nnesha

just do some example u will get the concept :=)

- 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

5. and then put the GCF outside

- Nnesha

right for perfect squares:
sign should be negative between terms
both numbers should be perfect squares
even power!

- calculusxy

i have a small question...
if we have -x^2, can we turn the x to be a positive ?

- calculusxy

...since the exponent is a positive, i thought that that's possible

- calculusxy

@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

and if we have \[\huge\rm -x^2+9\] take out the negative one
a should be positive

- calculusxy

-x^2 + y^3... i think this was a question on a test that i took

- 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

i moved -x^{-2} up to the numerator.
\[\frac{ -x^2 }{ y^3 } \implies \frac{ x^2 }{ y^3}\]

- calculusxy

would that be valid?

- Nnesha

ohhh no wait why did you flip the fraction ?

- Nnesha

IF IF `exponent ` is negative then you should flip the fraction

- calculusxy

i only moved -x^{-2} since the exponent was negative

- Nnesha

wait a sec its lagging let me refresh the page

- Nnesha

\[\huge\rm \frac{ 1 }{ -x^2y^3}\] is this ur question ?

- calculusxy

no\[\huge \frac{ 1 }{ -x^{-2}y^3 }\]

- Nnesha

ohh okay make sense

- 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

you forgot the negative sign
variables are correct :=)

- calculusxy

ok thank you!

- Nnesha

np :=) good luck!

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