
Synthetic division
Find the values of real numbers $\displaystyle a$ and $\displaystyle b$ such that:
$\displaystyle x^4 +4x^3+ax^2b$ is divisible by both $\displaystyle (x1)$ and $\displaystyle (x+2)$
Do I need to extract the factors (x1) and (x+2) FIRST and then try to solve for a and b or how do I start?
Also, the answers provided are a = 7 and b = 12 but when I try to divide $\displaystyle (x1)$ into $\displaystyle x^4+4x^3+7x^212$ I get a remainder of 1 which means that it does not divide it and a and b are incorrect values right?

Re: Synthetic division
If you make two horner scheme's, one with divisor 1 and the other one with divisor 2 (and at the end let the remainder =0) then you'll come to two simultaneous equations with variables a and b which you can solve directly.
(Notice: the coefficient of x is 0).

Re: Synthetic division
Ahh turns out I was just making errors in my synthetic division :S
Another method I used to get a = 7 and b = 12 is to just sub 1 and 2 into the original equation for x and solve these two equations simultaneously. Then I subbed these values in and divided by (x1) and then (x+2) to get zero remainder solutions.
Why does subbing in these values (1 and 2) for x allow you to find a and b? I can see that it works and I can see that they are derived from the (x1) and (x+2) terms but I don't see the logic behind it.
Thanks.

Re: Synthetic division
So have you tried it with the horner scheme?
1) divisor 1, the remainder becomes: $\displaystyle b+a+5$
And this remainder has to be equal to 0 if the polynomial is divisible by the factor (x1).
2) divisor 2, the remainder becomes: $\displaystyle b+4a16$
And this remainder has to be equal to 0 if the polynomial is divisible by the factor (x+2).
Therefore we get a system of simultaneous equations, if you multiply the second (or first) equation with a factor 1, then you can cancel out the variable b out and so find a and afterwards find b.

Re: Synthetic division
I am watching a video on it now :D I had never heard of it before.

Re: Synthetic division
But offcourse like you already said a more easy way is just entering the zero's into the equation and so you get the same system:
1) 1 is a zero, so: $\displaystyle 1+4+ab=0 \Leftrightarrow 5+ab=0$
2) 2 is a zero, so: $\displaystyle 1632+4ab=0 \Leftrightarrow 16+4ab=0$
So the same system :).