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 Dec 14, 2019
edited by whoisjoe  Jan 7, 2020
 #1
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Let the remainder be ax + b.  Then by the factor theorem, a = 4 and b = -1, so the remainder is 4x - 1.

 Dec 14, 2019
 #2
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Could someone answer this properly please?

 Dec 14, 2019
 #3
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On division by (x - 1)(x - 2) we would have

\(\displaystyle \frac{f(x)}{(x-1)(x-2)}=q(x)+\frac{r(x)}{(x-1)(x-2)},\)

where q(x) is of degree two less than f(x) and r(x) is linear,  ax + b say.

Multiply throughout by (x - 1)(x - 2) to get

\(\displaystyle f(x)=q(x)(x-1)(x-2)+ax + b\)

and now substitute x = 1 and x = 2.

That  gets you

\(\displaystyle f(1)=2=a+b,\\ f(2)=3=2a +b.\)

Solve those simultaneously to get a = 1 and b = 1.

 Dec 15, 2019
 #4
avatar+108790 
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Thanks guest.

I will take a proper look at your solution very soon.

Melody  Dec 15, 2019

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