# If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then relevant cubic has root between $0$ and $1$

If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then prove the function $$P(x)=a+bx+cx^2+dx^3$$ has a root somewhere between $0$ and $1$.

If it has a root between and $0$ and $1$ then I could show that it changes signs between $0$ and $1$. But some cubics have roots where they don't change signs.

$$P(1)=a+b+c+d\\P(0)=a$$ So $$P(1)=a+b+c+d=\\=a+b+c+d-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\=\frac{b}{2} + \frac{2c}{3} + \frac{3d}{4}\\P(0)=a=a-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\= -\frac{b}{2} -\frac{c}{3} - \frac{d}{4}$$ I feel like I'm close.

Although I'm sure there is a solution using the formula for cubic roots, I don't think I am supposed to use it, because the next part of the question (which I will hopefully be able to do myself) asks to generalize the result to any polynomial.

• Maybe the two expressions have opposite signs still? – Joao Noch Nov 7 '17 at 0:03
• Problem source is James Stewart Calculus 6E Ch5 (Integrals) – Joao Noch Nov 7 '17 at 0:04
• Because you need at least one coeff to be non zero you can get a root between 1/2 and 1. Because 1>1/3>1/4 and 1>1/4>1/8 then P(1)=a+b+c+d>a1+b2+c3+d4=0>a1+b2+c4+d8=P(1/2)P(1)=a+b+c+d>a1+b2+c3+d4=0>a1+b2+c4+d8=P(1/2) wait: i supposed they were positive, i am left with the rest still – user499752 Nov 7 '17 at 0:14

## 1 Answer

.Consider the polynomial defined by : $$Q(x) = ax + \frac{bx^2}{2} + \frac{cx^3}{3} + \frac{dx^4}{4}$$. Note that $$Q(0) = 0$$, since $$Q(x)$$ has constant term $$0$$. Furthermore, $$Q(1) = a + \frac b2 + \frac c3 + \frac d4 = 0$$. Therefore, we have located two roots of $$Q$$, $$0$$ and $$1$$.

Therefore, by Rolle's theorem, the derivative of $$Q$$ has a root between $$0$$ and $$1$$. But you can see that $$Q'(x) = P(x)$$, so that $$P(x)$$ has a root between $$0$$ and $$1$$.

• I'm liking it. That explains why it's in a calculus book (integral section) as well. – Joao Noch Nov 7 '17 at 0:14
• Thank you. This answer hit me when I saw the regularity of the expression $a + \frac b2 + \frac c3 + \frac d4 = 0$. – Teresa Lisbon Nov 7 '17 at 0:16