# $f(x)$ and $g(x)$ are monic cubic polynomials, with $f(x)-g(x)=r$. If $f$ has roots $r+1$ and $r+7$, and $g$ has roots $r+3$ and $r+9$, then find $r$.

Let $$f(x)$$ and $$g(x)$$ be two monic cubic polynomials, and let $$r$$ be a real number. Two of the roots of $$f(x)$$ are $$r+1$$ and $$r+7$$. Two of the roots of $$g(x)$$ are $$r + 3$$ and $$r + 9,$$ and$$f(x) - g(x) = r$$for all real numbers $$x.$$ Find $$r.$$

So far, I have $$f(x)=(x-r-1)(x-r-7)(x-p)$$ and $$g(x)=(x-r-3)(x-r-9)(x-q).$$ From $$f(x)-g(x)=r$$, I know that their constant terms differ by $$r$$. I expanded the two functions but it was too complicated. I also plugged in $$x=r+1,r+7,r+3,r+9$$ into $$f(x)-g(x)=r$$, but it didn't do much.

Hint: We have $$f(r+3)=r$$ (why ?) and $$p=\frac{9}{8}r + 3$$ easily follows. Then, $$f(r+9)=r=-2r + 96$$. Can you finish from there ?

Your approach is fine. Note that$$\begin{multline}f(x)-g(x)=\\=(4-p+q)x^2+(2 r p+8p-12 q-2 q r-4 r-20)x-p r^2+q r^2-8 p r+12 q r-7 p+27 q.\end{multline}$$So, $$4-p+q=0$$; in other words, $$p=q+4$$. Replacing $$p$$ with $$q+4$$ in the coefficient of $$x$$ in $$f(x)-g(x)$$, we get that $$4(3-q+r)=0$$; in other words, $$q=r+3$$. And if replace $$q$$ with $$r+3$$ in the constant term of $$f(x)-g(x)$$, we get $$32$$. But we want this to be equal to $$r$$. Therefore, $$r=32$$.

Some properties of these two polynomials may be worth remarking upon first:

• because $$\ f(x) \ = \ g(x) + r \ \ , \ \ r \$$ real, the curves of the two functions never intersect;

• for the same reason, the quadratic $$\ (b) \$$ and linear $$\ (c) \$$ coefficients are identical ;

• from the given information about their respective zeroes, we determine that $$g(r + 1) \ = \ -r \ \ ,$$ $$g(r + 3) \ = \ 0 \ \ , \ \ g(r + 7) \ = \ -r \ \ , \ \ g(r + 9) \ = \ 0 \ \ .$$

We will also use your notation to write $$\ f(p) \ = \ 0 \ \ , \ \ g(q) \ = \ 0 \ \ .$$ Since the polynomials have three real zeroes, their curves have two relative extrema (the cubic "S - curve"). This tells us something significant about the zeroes: "shifting" $$\ g(x) \$$ vertically by $$\ r \$$ moves the zeroes of $$\ g(x) \$$ at $$\ (r + 3) \$$ and $$\ (r + 9) \$$ "leftward" to the zeroes of $$\ f(x) \$$ at $$\ (r + 1) \$$ and $$\ (r + 7) \ \ ,$$ implying that $$\ r \ > \ 0 \ \ .$$ (The relative maximum moves away from the $$\ x-$$axis and the relative minimum, towards it.) The third zero of these polynomials must therefore be between the other two $$\ ( \ r + 1 \ \le \ p \ \le \ r + 7 \ \ , \ \ r + 3 \ \le \ q \ \le \ r + 9 \ ) \ \$$ and $$\ q \$$ should "move rightward" to $$\ p \ \ .$$ This last statement is confirmed from the Viete relations: the quadratic coefficient is $$b \ \ = \ \ -[ \ (r + 1) \ + \ p \ + \ (r + 7) \ ] \ \ = \ \ -[ \ (r + 3) \ + \ q \ + \ (r + 9) \ ]$$ $$\Rightarrow \ \ 2r \ + \ p \ + \ 8 \ \ = \ \ 2r \ + \ q \ + \ 12 \ \ \Rightarrow \ \ p \ = \ q + 4 \ \ .$$

For what follows, we will re-label the given zeroes in terms of $$\ \rho \ = \ (r + 3) \ \ . \$$ The linear coefficient of these polynomials is then $$c \ \ = \ \ (\rho - 2)·(q + 4) \ + \ (\rho + 4)·(q + 4) \ + \ (\rho - 2)·(\rho + 4)$$ $$= \ \ \rho · q \ + \ (\rho + 6) · q \ + \ \rho · (\rho + 6)$$ $$\Rightarrow \ \ \rho^2 \ + \ 2·q·\rho \ + \ 10·\rho \ + \ 2·q \ \ = \ \ \rho^2 \ + \ 2·q·\rho \ + \ 6·\rho \ + \ 6·q \ \ \Rightarrow \ \ \rho \ \ = \ \ q \ \ .$$ So we discover that $$\ q \ = \ (r + 3) \$$ is in fact a double zero of $$\ g(x) \ \ .$$ In turn, we find that $$\ p \ = \ q + 4 \ = \ (r + 3) + 4 \ = \ (r + 7) \$$ is a double zero of $$\ f(x) \ \ .$$

If we designate $$\ d \$$ as the "constant term" of $$\ g(x) \ \ , \$$ we find $$d \ = \ -(r + 3)^2·(r + 9) \ \ \Rightarrow \ \ d \ + \ r \ = \ r \ - \ (r + 3)^2·(r + 9) \ \ = \ \ -(r + 1)·(r + 7)^2$$ $$\Rightarrow \ \ -r^3 \ - \ 15r^2 \ - \ 62r \ - \ 81 \ \ = \ \ -r^3 \ - \ 15r^2 \ - \ 63r \ - \ 49 \ \ \Rightarrow \ \ r \ = \ 32 \ \ .$$

Our polynomials are therefore $$g(x) \ \ = \ \ (x - 35)^2 \ · \ (x - 41) \ \ = \ \ x^3 \ - \ 111x^2 \ + \ 4095x \ - \ 50225 \ \ \ \text{and}$$

$$f(x) \ \ = \ \ (x - 33) \ · \ (x - 39)^2 \ \ = \ \ x^3 \ - \ 111x^2 \ + \ 4095x \ - \ 50193 \ \ = \ \ g(x) \ + \ 32 \ \ .$$ [The relative extrema of $$\ g(x) \$$ are located at $$\ (35 \ , \ 0) \$$ and $$\ (39 \ , \ -32) \ \ ,$$ while those of $$\ f(x) \$$ are $$\ (35 \ , \ 32) \$$ and $$\ (39 \ , \ 0) \ \ . \ ]$$

Incidentally, there is a complementary pair of polynomials for $$\ r \ = \ -32 \ \ :$$ $$g(x) \ \ = \ \ (x + 25)^2 \ · \ (x + 31) \ \ = \ \ x^3 \ + \ 81x^2 \ + \ 2175x \ + \ 19375 \ \ \ \text{and}$$ $$f(x) \ \ = \ \ (x + 23) \ · \ (x + 29)^2 \ \ = \ \ x^3 \ + \ 81x^2 \ + \ 2175x \ + \ 19343 \ \ = \ \ g(x) \ - \ 32 \ \ .$$

Let $$\alpha$$, $$\beta$$, and $$\gamma$$ be the roots of a monic cubic polynomial $$p$$. Then:

$$p(x) = (x - \alpha)(x - \beta)(x - \gamma)$$ $$= x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \alpha\gamma + \beta\gamma)x - \alpha\beta\gamma$$

Matching up the coefficients with the general cubic $$x^3 + bx^2 + cx + d$$ (I'm omitting $$a$$ because we're given that it's 1) gives:

$$\alpha + \beta + \gamma = -b$$ $$\alpha\beta + \alpha\gamma + \beta\gamma = c$$ $$\alpha\beta\gamma = -d$$

These equations are called Vieta's formulas.

Now, let's consider two polynomials that are simple horizontal shifts of the given ones:

$$f_r(x) = f(x-r): \alpha = 1, \beta = 7, \gamma = s$$ $$g_r(x) = g(x-r): \alpha = 3, \beta = 9, \gamma = t$$

Since they differ only by a constant ($$r$$), they must have the same $$b$$ and $$c$$ coefficients, and have $$d$$ differ by $$r$$. So,

$$1 + 7 + s = 3 + 9 + t$$ $$7 + s + 7s = 27 + 3t + 9t$$ $$-7s + 27t = r$$

Or, rearranging a bit:

$$s - t = 4\tag{1}$$ $$8s - 12t = 20\tag{2}$$ $$r + 7s - 27t = 0\tag{3}$$

A simple linear system of equations that can be solved for $$r$$, $$s$$, and $$t$$. From (1), $$s = t + 4$$. Substituting this into (2) gives:

$$8(t + 4) - 12t = 20$$ $$8t + 32 - 12t = 20$$ $$-4t = -12$$ $$t = 3$$ $$s = t + 4 = 7$$

Finally, substituting into (3) gives:

$$r + 7(7) - 27(3) = 0$$ $$r + 49 - 81 = 0$$ $$r = 32$$