# From the given information, how do we use the Remainder Theorem to reach the answer?

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 $$10$$.

Use the Remainder Theorem. We have

$$-a+ar-ar^2+ar^3=-30$$

and

$$a+ar+ar^2+ar^3=10$$

Solving simultaneously:

$$\frac{a(1-r^4)}{1+r}=30$$

and

$$\frac{a(1-r^4)}{1-r}=10$$

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

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

• 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. – Arturo Magidin Mar 26 at 1:16
• OK, noted, thanks. – umzung Mar 26 at 1:18
• +1 for taking the road to mathjax. – 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 $$-1$$.