Newton's Sums

Algebra Level 1

$P(x)=x^{3}+33x^{2}+327x+935$

Let $P(x)$ be a polynomial as described above with $a, b, c$ the roots of $P(x)$ .

Find $a^{2}+b^{2}+c^{2}$ without solving $P(x)=0$ .

The answer is 435.

2 solutions

William Isoroku
Dec 26, 2014

It is helpful to know that ${a}^{2}+{b}^{2}+{c}^{2}=(a+b+c)^{2}-2(ab+bc+ac$

Then apply Vieta's formulas:

$(a+b+c)^{2}=(-\frac{coefficient\:of\: the\: second\: highest\: power} {leading\: coefficient})^{2}$ which is $(\frac{-33} {1})^{2}=33^2$

$2(ab+bc+ac)=2(\frac{coefficient\:of\:third\:highest\:power} {leading\:coefficient})=2(327)$

Substitution: ${33}^{2}-2(327)=\boxed{435}$

Moderator note:

A nice standard approach.

Bonus question : Find $a^{2}+b^{2}+c^{2}$ by solving $P(x)=0$ instead.

The first statement of your solution is one of the Newton's identities for the power sums of the roots of a polynomial $P(x)$ , which is,

$P_2=e_1P_1-2e_2$ , where $e_k$ refers to the $k^{\text{th}}$ elementary symmetric polynomial and $P_k$ refers to the $k^{\text{th}}$ power sum of the roots. Rest of the work is simply done using Vieta's formulas to find out $e_1,e_2$ and then solve for $P_2$ . Also, note that $P_1=e_1$ .

Prasun Biswas - 6 years, 5 months ago

I used the same method!

Swapnil Das - 6 years ago

If we solve P(x) = 0 we get the roots as -5 , -11 , and -17 So The required sum is 289+121+25 = 435

Sahil Sharma - 3 years, 8 months ago

Eliott Collin
Dec 25, 2014

As suggested in the title of the problem, you must be familiar to Newton's Sums. Knowing that $P(x)=x^{3}+33x^{2}+327x+935$ , we can affirm that:

$a+b+c+33=0$

$a^{2}+b^{2}+c^{2}+33(a+b+c)+2*327=0$

Clearly, $a^{2}+b^{2}+c^{2}=435$

Learn more on Newton's Sums here : http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums

Moderator note:

You don't actually need to heed the title's suggestion as not all of them are accurately. Do you really need to solve this by Newton's Sum? Read Prasun Biswas' comment under William Isoroku's solution or read Anuj Modi's comment.

Elliot, can you explain how you found 435?

Emmanuel Ajimoko - 6 years, 5 months ago

33 = a + b + c as (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ac)x + abc 1089 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac) ab+bc+ac = 327 then substitute and subtract

Anuj Modi - 6 years, 5 months ago

Newton's Sums

The answer is 435.

2 solutions

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