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how can i factor 10t2−33t+26 completely
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check the question again @michaelshaw
it says "factor completely" 10t2−33t+26
@mathslover
oh k I got it now
\[\large{10t^2 -20t -13t +26}\] \[\large{10t(t-2)-13t(t-2)}\]
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can u tell me what will we get after taking (t-2) common?
(10t-13)(t-2) ?
great work so that is ur answer
thanks man!
@mathslover
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i have another one if ur up for it..
No problem also: \[\huge{\mathbb{WELCOME}\space \textbf{TO}\space \cal{OpenStudy}}\]
33t2+13t−6 AND THANKS I LIKE IT HERE!
33t^2+13t-6
@mathslover
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please post this as a new question it will be answered there
ok
i dont know how to, did i just do it?
yes
thanks
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