How do you simplify #(6t^3 + 5t^2 + 9) / (2t + 3)#?

2 Answers
Kigis
May 10, 2018

#3t^2 - 2t^2 + 3 #

Explanation:

#(6t^3 + 5t^2 + 9) / (2t+3)#

#(2t+3(3t^2-2t+3)) / (2t+3)#

  • 2t+3 on numerator and denomoniator cancel out.

ANSWER: #3t^2 - 2t^2 + 3 #

You have to factorise the top that #2t + 3# directly factorises #6t^3 + 5t^2 + 9# so that it cancels the denominator giving you a value of #3t^2 - 2t^2 + 3#.

The are many ways of factorising #6t^3 + 5t^2 + 9#, polynomial division, rational root theorem.

I would recommend polynomial division as its pretty straight forward to learn and rely on. Here is a link that will teach you all about it and you can practise with: http://www.purplemath.com/modules/polydiv2.htm

Jim G.
May 10, 2018

#3t^2-2t+3#

Explanation:

#"one way is to use the divisor as a factor in the numerator"#

#"consider the numerator"#

#color(red)(3t^2)(2t+3)color(magenta)(-9t^2)+5t^2+9#

#=color(red)(3t^2)(2t+3)color(red)(-2t)(2t+3)color(magenta)(+6t)+9#

#=color(red)(3t^2)(2t+3)color(red)(-2t)(2t+3)color(red)(+3)(2t+3)cancel(color(magenta)(-9))cancel(+9)#

#=color(red)(3t^2)(2t+3)color(red)(-2t)(2t+3)color(red)(+3)(2t+3)+0#

#rArr(cancel((2t+3))(3t^2-2t+3))/cancel((2t+3))=3t^2-2t+3#

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