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Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

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 $324-80$=$244$

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