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Mathematics 16 Online
OpenStudy (anonymous):

a2-2a2-35

ganeshie8 (ganeshie8):

it must be : \(\large a^2 - 2a - 35\) ?

OpenStudy (anonymous):

the 2a is squared

OpenStudy (anonymous):

im sorry i typed the problem wrong the first a is to the 4th

ganeshie8 (ganeshie8):

\(\large a^4 - 2a^2 - 35\)

ganeshie8 (ganeshie8):

like that ?

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

and you wanto factor it ?

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

\(\large a^4 - 2a^2 - 35\) say \(a^2 = x\), then the expression becomes : \(\large x^2 - 2x - 35\)

ganeshie8 (ganeshie8):

can u factor that ?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

i dont think so anyway

ganeshie8 (ganeshie8):

\(\large x^2 - 2x - 35\)

ganeshie8 (ganeshie8):

find two numbers such that : product = -35 sum = -2

OpenStudy (anonymous):

-5 & 7?

OpenStudy (anonymous):

but then I get a +2 instead of -2

ganeshie8 (ganeshie8):

try -7 & 5

OpenStudy (anonymous):

oh duh!

OpenStudy (anonymous):

but is that with the second part being 2a2

ganeshie8 (ganeshie8):

yes :) so whats the factored form ?

ganeshie8 (ganeshie8):

yup

ganeshie8 (ganeshie8):

\(\large x^2 - 2x - 35\) factors to \(\large (x-7)(x+5)\)

ganeshie8 (ganeshie8):

plugin x = a^2 above

ganeshie8 (ganeshie8):

\(\large (x-7)(x+5)\) \(\large (a^2-7)(a^2+5)\)

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