Synthetic division of polynomials by factor of the form $(ax+b)$
Find the quotient and remainder when $6x^4-11x^3+5x^2-7x+9$ is divided by $(2x-3)$. I expressed the divisor $(2x-3)$ as $2(x-\frac{3}{2})$ and conducted synthetic division by $\frac{3}{2}$ and obtained the coefficients of the quotients as $6, -2, 2$ and $-4$ and the remainder as $3$. Then I divided the quotients by $2$. Why should I do this? and why should I not divide the remainder by $2$?
$\endgroup$2 Answers
$\begingroup$Synthetic division is simply a shortcut to polynomial long division. But it only works for monic linear factors, i.e. those of the form $(x-\alpha)$.
When you divide a polynomial $P(x)$ by a non-monic linear factor like $(ax-b), a \neq 1$, this is what you get:
$$P(x) = Q(x)(ax-b) + R$$
where $Q(x)$ is the quotient polynomial and $R$ is the remainder (it will be a constant term).
That can also be expressed as :
$$P(x) = aQ(x)(x-\frac{b}{a}) + R$$
Note that when you're doing synthetic division by $\displaystyle (x-\frac{b}{a})$, you're actually getting $aQ(x)$ as the quotient. Hence the quotient $Q(x)$when you divide by $(ax-b)$ has to be determined by dividing $aQ(x)$ by $a$. That's why you're doing that division step. Note also that the remainder $R$ is unaltered, which is why you don't have to do anything to it.
In your specific example, $a=2, b = 3$. You're doing the synthetic division by $\displaystyle (x-\frac{3}{2})$ and I hope it should now be clear why you need to divide the quotient you get by $2$ to get the quotient when you divide by $(2x-3)$.
$\endgroup$ $\begingroup$In my opinion, you do not need to change $(2x-3)$ by anything. If you perform the long divition of $6x^4-11x^3+5x^2-7x+9$ by $(2x-3)$, you should directly obtain $$\frac{6x^4-11x^3+5x^2-7x+9 }{2x-3}=3 x^3-x^2+x-2+\frac{3}{2 x-3}$$ If i May suggest, don't make your life more complex than it is. Cheers.
$\endgroup$ 1
