+0

# /gone

-1
67
4
+43

/gone

Dec 14, 2019
edited by whoisjoe  Jan 7, 2020

#1
0

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
+106915
0

Dec 14, 2019
#3
+1

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
+106915
0

Thanks guest.

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

Melody  Dec 15, 2019