How do you divide #(6x^2-31x+5) / (x-5)#?

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Answer 1 · 2016-11-16T08:26:02

The answer is #=(6x-1)#

Let #f(x)=6x^2-31x+5#

Then, #f(5)=6*25-31*5+5=150-155+5=0#

The remainder is #0#, so #f(x)# is divisible by #(x-5)#

Let's do a long division

#color(white)(aaaa)##6x^2-31x+5##color(white)(aaaa)##∣##x-5#

#color(white)(aaaa)##6x^2-30x##color(white)(aaaaaaaa)##∣##6x-1#

#color(white)(aaaaaa)##0-x+5#

#color(white)(aaaaaaaa)##-x+5#

#color(white)(aaaaaaaa)##-0+0#

Now, we can factorise the numerator

#6x^2-31x+5=(6x-1)(x-5)#

Then,

#(6x^2-31x+5)/(x-5)=((6x-1)cancel(x-5))/cancel(x-5)=(6x-1)#