Try this beautiful Problem on Algebra based on Roots of Polynomial from AMC 10 A, 2019. You may use sequential hints to solve the problem.

## Algebra- AMC-10A, 2019- Problem 24

Let $p, q,$ and $r$ be the distinct roots of the polynomial $x^{3}-22 x^{2}+80 x-67$. It is given that there exist real numbers $A, B$, and $C$ such that $\frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r}$

for all $s \notin{p, q, r} .$ What is $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} ?$

,

- $243$
- $244$
- $245$
- $246$
- $247$

**Key Concepts**

Algebra

Linear Equation

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2019 Problem-24

#### Check the answer here, but try the problem first

$244$

## Try with Hints

#### First Hint

The given equation is $\frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r}$…………………(1)

If we multiply both sides we will get

Multiplying both sides by $(s-p)(s-q)(s-r)$ we will get

$$

1=A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q)

$$

Now can you finish the problem?

#### Second Hint

Now Put $S=P$ we will get $\frac{1}{A}=(p-q)(p-r)$…………(2)

Now Put $S=q$ we will get $\frac{1}{B}=(q-p)(q-r)$………..(3)

Now Put $S=r$ we will get $\frac{1}{C}=(r-p)(r-q)$………..(4)

Now Can you finish the Problem?

#### Third Hint

Adding (2) +(3)+(4) we get,$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^{2}+q^{2}+r^{2}-p q-q r-p r$

Now Using Vieta’s Formulas, $p^{2}+q^{2}+r^{2}=(p+q+r)^{2}-2(p q+q r+p r)=324$ and $p q+q r+p r=80$

Therefore the required answer is $324-80$=$244$

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=h4MmDUky7KM