Why does the sequence for the polynomial alternate when it equals $-30$, and not alternate when it equals $10$? Why is the denominator when the sum of the geometric series is equal to $30$, $1+r$, and not $1-r$? I am unable to correctly relate the Remainder Theorem to its use in the given answer.

The question:

When a polynomial of degree $3$ is written in descending powers of $x$ the coefficients of its four terms form a geometric sequence.

When the polynomial is divided by $x+1$ it s remainder is $-30$. Then it is divided by $x-1$ its remainder is $10$.

The answer:

Use the Remainder Theorem. We have




Solving simultaneously:




Giving $r=-\frac{1}{2}$ and $a=16$

The polynomial is therefore $16x^3-8x^2+4x-2$

  • $\begingroup$ I know this would mean an lot of typing, but still: please do not rely on images for the content of your post. They are not searchable, and often screen readers cannot handle them. $\endgroup$ – Arturo Magidin Mar 26 at 1:16
  • $\begingroup$ OK, noted, thanks. $\endgroup$ – umzung Mar 26 at 1:18
  • $\begingroup$ +1 for taking the road to mathjax. $\endgroup$ – Paramanand Singh Mar 26 at 3:49

The polynomial is a cubic whose coefficients are a geometric sequence (in the order of descending powers, i.e. starting with $x^3$ and ending with the constant term), meaning that our polynomial must take the form $$p(x) = ax^3 + arx^2 + ar^2x + ar^3.$$ Remainder theorem states that the remainder of $p(x)$ when divided by $x - \alpha$, is $p(\alpha)$.

So, when dividing $p(x)$ by $x - 1$ (i.e. $\alpha = 1$), the remainder is $p(1)$. Thus, $$10 = p(1) = a1^3 + ar1^2 + ar^21 + ar^3 = a + ar + ar^2 + ar^3.$$

Similarly, when dividing $p(x)$ by $x + 1$ (i.e. $\alpha = -1$), the remainder is $p(-1)$. Thus, $$-30 = p(-1) = a(-1)^3 + ar(-1)^2 + ar^2(-1) + ar^3 = -a + ar - ar^2 + ar^3.$$

This should hopefully explain the alternating signs. The next observation you need to make is that both of the above expressions are geometric series. Each have $4$ terms. The first has an initial term of $a$ and a common ratio of $r$, while the second has an initial term of $-a$ and a common ratio of $-r$. The first one is precisely the usual formula for geoemetric series: $$10 = a + ar + ar^2 + ar^3 = a\frac{1 - r^4}{1 - r}.$$

To get a formula for the second one, we simply replace $r$ with $-r$ and $a$ with $-a$, to get $$-30 = -a + ar - ar^2 + ar^3 = (-a)\frac{1 - (-r)^4}{1 - (-r)} = -a\frac{1 - r^4}{1 + r}.$$ To get the equation from the solution, simply multiply both sides by $-1$.

  • $\begingroup$ Thanks, that's very clear @Theo Bendit $\endgroup$ – umzung Mar 26 at 1:45
  • $\begingroup$ +1 for very clear explanation. $\endgroup$ – Paramanand Singh Mar 26 at 3:50

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