How do you simplify #(t^5 - 3t^2 - 20) (t-2)^-1#?

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Answer 1 · 2018-06-25T08:45:57

#(t^5-3t^2-20)(t-2)^(-1)=t^4+2t^3+4t^2+5t+10#

We can write #(t^5-3t^2-20)(t-2)^(-1)# as

#(t^5-3t^2-20)/(t-2)#

Now observe that for the numerator #f(t)=t^5-3t^2-20#,

#f(2)=2^5-3*2^2-20=32-12-20=0# and from factor theorem #(t-2)# is a factor of #t^5-3t^2-20#

and hence we can divide #t^5-3t^2-20# by #t-2# and

#t^5-3t^2-20#

= #t^4(t-2)+2t^3(t-2)+4t^2(t-2)+5t(t-2)+10(t-2)#

= #(t-2)(t^4+2t^3+4t^2+5t+10#

and hence #(t^5-3t^2-20)(t-2)^(-1)=t^4+2t^3+4t^2+5t+10#