Thread:

1. Synthetic division

Find the values of real numbers $a$ and $b$ such that:

$x^4 +4x^3+ax^2-b$ is divisible by both $(x-1)$ and $(x+2)$

Do I need to extract the factors (x-1) 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 $(x-1)$ into $x^4+4x^3+7x^2-12$ I get a remainder of 1 which means that it does not divide it and a and b are incorrect values right?

2. 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).

3. 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 (x-1) 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 (x-1) and (x+2) terms but I don't see the logic behind it.

Thanks.

4. Re: Synthetic division

So have you tried it with the horner scheme?
1) divisor 1, the remainder becomes: $-b+a+5$
And this remainder has to be equal to 0 if the polynomial is divisible by the factor (x-1).
2) divisor -2, the remainder becomes: $-b+4a-16$
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.

5. Re: Synthetic division

I am watching a video on it now I had never heard of it before.

6. 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: $1+4+a-b=0 \Leftrightarrow 5+a-b=0$
2) -2 is a zero, so: $16-32+4a-b=0 \Leftrightarrow -16+4a-b=0$

So the same system .
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