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Algebra 21 Online
OpenStudy (elleblythe):

Step-by-step solution in how to factor x^4-7x^2y^2+9y^2

Directrix (directrix):

Question: Can you factor a^2 - 7a + 9 ? If so, then the posted expression will factor? If not, then the posted expression will not factor. @elleblythe

OpenStudy (mayankdevnani):

\[\huge x^4-7x^2y^2+9y^2\] \[\large there~ are~ two~factors \] \[\large one~factor:-\] \[\huge x^2(x^2-7y^2)+9y^2\] \[\large second~factor:-\] \[\huge x^4+y^2(9-7x^2) ~~~~OR~~~ x^4+(9-7x^2)y^2\]

Directrix (directrix):

Those are not factors.

OpenStudy (elleblythe):

@Directrix @mayankdevnani the answer is supposed to be (x^2-xy-3y^2)(x^2+xy-3y^2) but what's the solution?

OpenStudy (mayankdevnani):

@Directrix thank you.... now i can factor this.. please wait

ganeshie8 (ganeshie8):

ok, then the question has a typo

ganeshie8 (ganeshie8):

check if the given expression is :- \(x^4-7x^2y^2+9\color{red}{y^4}\)

OpenStudy (elleblythe):

sorry my bad! the given expression is x^4−7x^2y^2+9y^4. thanks for correcting me @ganeshie8

ganeshie8 (ganeshie8):

\(x^4-7x^2y^2+9y^4\) say, \(x^2=m, y^2=n\) \(m^2-7mn+9n^2\) \((m-3n)^2 -mn\) \((x^2-3y^2)^2 - x^2y^2\) \((x^2-3y^2+xy)(x^2-3y^2-xy)\)

ganeshie8 (ganeshie8):

see if that makes some sense

OpenStudy (mayankdevnani):

\[\huge x^4 - 7x^2y^2 + 9y^4\] \[\large we~can~write~It~also~as:-\] \[\huge x^4-6x^2y^2-x^2y^2+9y^4\] \[\huge ~~~~~~~~~~~OR~~~~~~~~\] \[\huge x^4-6x^2y^2+9y^4-x^2y^2\] \[\huge (x^2-3y^2)^2-(xy)^2\] \[\huge (x^2-3y^2+xy)(x^2-3y^2-xy)\]

OpenStudy (mayankdevnani):

understood or not? @elleblythe

OpenStudy (elleblythe):

got it thank you @ganeshie8 @mayankdevnani

OpenStudy (mayankdevnani):

welcome :)

ganeshie8 (ganeshie8):

np :) your questions are very tough :S

OpenStudy (jhannybean):

\begin{align} \large \color{blue}{x^4}-7\color{green}{x^2y^2}+\color{blue}{9y^4} &= \large x^4 -6x^2y^2-x^2y^2 +9y^4 \\ &=\large \color{blue}{ x^4 -6x^2y^2+9y^4}-\color{green}{x^2y^2} \\ &= \large (x^2-3y^2)^2 -(xy)^2 \\ &=\large \color{blue}{(x^2 -3y^2)}^2 -\color{green}{(x^2y^2)} \\ &=\large (x^2-3y^2 +x^2y^2)(x^2-3y^2 -x^2y^2) \\ \end{align}

ganeshie8 (ganeshie8):

wow its nice ! where did u learn align tag @Jhannybean

OpenStudy (jhannybean):

By creeping on people's TeX formatting LOL

ganeshie8 (ganeshie8):

thats fast way to learn latex :) how do we control wat it aligns wid ?

OpenStudy (jhannybean):

&=

ganeshie8 (ganeshie8):

\( \begin{align} ganeshie &= g \\ &= a \\ &= n \\ &= e\\ &= s\\ &= h\\ \end{align} \)

OpenStudy (jhannybean):

oh i forgot the \\ ahah.

ganeshie8 (ganeshie8):

\( \begin{align} Jhanny &bean \\ & J \\ & h \\ & a\\ & n\\ & n\\ & y\\ \end{align} \)

ganeshie8 (ganeshie8):

aligned it to b

ganeshie8 (ganeshie8):

basically we can align it to watever we want, just put & before watever u want to align the next line to, nice :) feels like an expert lol

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