No bash please

Algebra Level 5

$\large P(x)=x^4+7x^3+2x-9$

If $a_1,a_2,a_3,a_4$ are roots of $P(x)=0$ , find the sum of coefficients of a least degree monic polynomial whose roots are $a_1a_2a_3,a_1a_2a_4,a_1a_3a_4,a_2a_3a_4$ .

The answer is -159.

2 solutions

Nihar Mahajan
Jun 16, 2015

Let the roots be written in form $\dfrac{\displaystyle\prod_{i=1}^4 a_i}{a_i} = \dfrac{-9}{a_i}$

We introduce $t_i$ which is root of our required equation such that:

$t_i=\dfrac{-9}{a_i} \quad \Rightarrow a_i=\dfrac{-9}{t_i} \quad \forall \ i \ 1\leq i \leq 4$

Then we have:

$\forall \ i \ 1\leq i \leq 4\begin{cases}{\begin{array}{c}(P\left(\dfrac{-9}{t_i}\right) = 0 \\ \Rightarrow \left(\dfrac{-9}{t_i}\right)^4+7\left(\dfrac{-9}{t_i}\right)^3+2\left(\dfrac{-9}{t_i}\right)-9=0 \\ \Rightarrow \dfrac{9^4}{t_i^4}+\dfrac{-9^3\times 7}{t_i^3}+\dfrac{-9\times 2}{t_i}-9=0 \\ \Rightarrow \dfrac{-t_i^4}{9}\left(\dfrac{9^4}{t_i^4}+\dfrac{-9^3\times 7}{t_i^3}+\dfrac{-9\times 2}{t_i}-9\right)=0 \\ \Rightarrow t_i^4+2t_i^3+567t_i-729=0\end{array}}\end{cases}$

Thus sum of coefficients $=1+2+567-729=\huge\boxed{-159}$

You don't need to do that much work (expanding out the new polynomial). By definition of a polynomial, if we have a polynomial $p$ , the sum of coefficients of all terms of $p$ is equal to $p(1)$ . We exploit this idea to get our answer.

Denote by $e_i$ the $i^{\textrm{th}}$ elementary symmetric polynomial made up by the roots of $P(x)$ . By Vieta's formulas, we have,

$e_1=(-7)~,~e_2=0~,~e_3=(-2)~,~e_4=(-9)$

As you noticed, the new four roots in the problem are $e_4/a_i$ where $1\leq i\leq 4$ . As such, our new polynomial (call it $g$ ) can be written as follows (using Remainder-Factor Theorem):

$g(x)=\prod_{k=1}^4\left(x+\frac 9{a_k}\right)\implies g(1)=\prod_{k=1}^4\left(1+\frac 9{a_k}\right)\\ \implies g(1)=\frac{\displaystyle\prod_{k=1}^4(a_k+9)}{\displaystyle\prod_{k=1}^4 a_k}=\frac{e_4+9e_3+9^2e_2+9^3e_1+9^4}{e_4}$

Plugging values, we get $g(1)=(-159)$ which is our answer.

Prasun Biswas - 5 years, 12 months ago

Good approach. Note that we don't even need to know what $e_i$ are. The answer is equal to $\frac{ f( -9) } { f(0) }$ .

Do you see why?

Calvin Lin Staff - 5 years, 12 months ago

I guess you meant $P(x)$ instead of $f(x)$ . And the reason is simple. As a consequence of Vieta's Formulas, we can write $P(x)$ as,

$P(x)=e_0x^4-e_1x^3+e_2x^2-e_3x+e_4\ \ \ \ \ (i)$

In the expression for $g(1)$ , the numerator matches $(i)$ with $x=(-9)$ and the denominator matches $(i)$ with $x=0$ . Hence, the answer is $\frac{P(-9)}{P(0)}$ .

Note: $~~~e_0=1$ by definition.

I wonder why my previous comment got downvoted 2 times. Makes me sad. :\

EDIT: One of the downvoters suddenly removed his/her downvote. This is becoming weird, to be honest.

Prasun Biswas - 5 years, 12 months ago

Yaar...i jus missed it...sirf answer wrong aaya!!!! Bye the way u r in which class???

sarvesh dubey - 5 years, 12 months ago

I am in class 10.