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Please help me to solve this problem using Ruffini's rule.

Given $P(x) = a x^3 + 2 x^2 + c x + d$, please help me to determine the values of $a$, $c$ and $d$ so that P(x) is divisible by $(3x+2)$ and by $(x+2)$.

Stefan4024
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UltraCommit
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1 Answers1

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As lab bhattacharjee mentioned it's best to use Polynomial remainder theorem. It states that $x-a$ is divisor of $f(x)$, iff $f(a) = 0$.

In this case we have 2 divisors, which are $(3x+2)$ and $(x+2)$. So for the first divisor $a=- \frac 23$ and for the second $a= -2$

Now we plug them in the function:

$$f(-\frac 23) = a\left(-\frac 23\right)^3 + 2\left(-\frac 23\right)^2 - c\frac 23 + d = 0$$

$$- \frac{8a}{27} + \frac 89 - \frac{2c}{3} + d = 0$$

$$f(-2) = a(-2)^3 + 2(-2)^2 - 2c + d = 0$$

$-8a + 8 - 2c + d = 0$

Solving these two linear equation it gives that any ordered triple:

$$(a,c,d) = (-\frac{9t-24}{32}, \frac{13t + 8}{8}, t); \forall t \in \mathbb{R}$$

If you are interested in just integer solution we obtain: $t \equiv 0 (\mod 8)$ from the form for $c$ and $t \equiv 24 (\mod 32)$. This leads to conclusion that t is of the form $32k + 24$

So for integer solution we have:

$$(a,b,c) = (-9k-6, 52k + 40,32k + 24); \forall k \in \mathbb{R}$$

Stefan4024
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  • Thank you very much for your detailed explanation. My brother's book shows only the solution a = 3 AND c = -12 AND d = -8. I don't know why. There are infinite solutions! – UltraCommit Sep 17 '13 at 21:20
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    Maybe it's just showing one possible solution. – Stefan4024 Sep 17 '13 at 21:24
  • You're right, but I think that this approach is not the best for explaining the concept to children. The book's author has implicitly selected one solution formed by only integer coefficients, although this was not explicitly required. Thank you again for having helped me to understand. – UltraCommit Sep 17 '13 at 21:28