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 Skip Navigation LinksMath Help > Basic Algebra > Factoring > Solving cubic, quartic, higher degree > Cubic polynomial puzzle

On 9/7/01 12:55:21 AM, Jagjot Singh wrote:
>2) Suppose the roots of
>x³ + 3x² + 4x - 11 = 0
>are a, b and c and the roots of
>x³+rx²+sx+t=0 are
>a+b, b+c, and a+c. Find the
>value of t.

(x-a)(x-b)(x-c) = x³ + 3x² + 4x - 11

x³ - (a+b+c)x² + (ab+ac+bc)x - abc = x³ + 3x² + 4x - 11

Now by equating coefficients, we know the following things:

a+b+c=-3
ab+ac+bc=4
abc=11

I will use these facts, when needed, for substitutions. Now for the second equation:

(x-a-b)(x-b-c)(x-a-c) = x³+rx²+sx+t

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ac)x - (a+b)(b+c)(a+c) = x³+rx²+sx+t

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ac)x + (3+a)(3+b)(3+c) = x³+rx²+sx+t

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ac)x + (27+3ab+3ac+3bc+9a+9b+9c+abc) = x³+rx²+sx+t

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ac)x + (27+3*4-9*3+11) = x³+rx²+sx+t

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ac)x + 23 = x³+rx²+sx+t

So t=23. That answers the question. But let's not stop, we're on a roll!

First,
a²+b²+c²=1,
because a²+b²+c² = (a+b+c)²-2(ab+ac+bc) = (-3)²-2(4) = 1

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ac)x + 23 = x³+rx²+sx+t

x³ - 2(-3)x² + (1+3*4)x + 23 = x³+rx²+sx+t

x³ + 6x² + 13x + 23 = x³+rx²+sx+t

So r=6, s=13, and t=23.

I won't bore you with the details, but I checked that this solution is right by using a computer to solve the two cubics, and verifying the roots of the second one are a+b, a+c, and b+c, where a, b, and c are the roots of the first one.


The method I used to answer this question suggests that a more general case can be solved...

Show that if the roots of

x³ + dx² + ex + f = 0

are a, b and c then the roots of

x³ + (2d)x² + (d²+e)x + (de-f) = 0

are a+b, b+c, and c+a.

(x-a)(x-b)(x-c) = x³ + dx² + ex + f

x³ - (a+b+c)x² + (ab+ac+bc)x - abc = x³ + dx² + ex +f

Now, by equating coefficients, we know the following things:

a+b+c = -d
ab+bc+ca = e
abc = -f

Also,

a²+b²+c² = d²-2e,
   because a²+b²+c² = (a+b+c)²-2(ab+bc+ca)

I will use these facts, when needed, for substitutions.

(x-a-b)(x-b-c)(x-c-a) =

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ca)x - (a+b)(b+c)(c+a) =

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ca)x + (d+a)(d+b)(d+c) =

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ca)x + (d³+d²(a+b+c)+d(ab+bc+ca)+abc) =

x³ - 2(a+b+c)x² + (a²+b²+c²+3ab+3bc+3ca)x + (de-f) =

x³ - 2(a+b+c)x² + (d²-2e+3e)x + (de-f) =

x³ - 2(a+b+c)x² + (d²+e)x + (de-f) =

x³ + (2d)x² + (d²+e)x + (de-f)


One more similar problem, for good measure...

Show that if the roots of

x³ + dx² + ex + f = 0

are a, b and c then the roots of

x³ + (?)x² + (?)x + (?) = 0

are a-b, b-c, and c-a.

(x-a)(x-b)(x-c) = x³ + dx² + ex + f

x³ - (a+b+c)x² + (ab+ac+bc)x - abc = x³ + dx² + ex +f

So we still know the following things, by equating coefficients as before:

a+b+c = -d
ab+bc+ca = e
abc = -f

Also,

a²+b²+c² = d²-2e,
   because a²+b²+c² = (a+b+c)²-2(ab+bc+ca)

I will use these facts, when needed, for substitutions.

(x-a+b)(x-b+c)(x-c+a) =

x³ + (-a²-b²-c²+ab+bc+ca)x - (a-b)(b-c)(c-a) =

x³ + (-d²+2e+e)x - (a-b)(b-c)(c-a) =

x³ + (-d²+3e)x - (a-b)(b-c)(c-a) =

x³ + (-d²+3e)x - (ab²-ac²+bc²-ba²+ca²-cb²) =

x³ + (-d²+3e)x - (a²(c-b)+b²(a-c)+c²(b-a)) =

x³ + (-d²+3e)x + (d+c+2b)(d+a+2c)(d+b+2a) =

x³ + (-d²+3e)x + (d³ + (3d²)(a+b+c) + (2d)(a²+b²+c²) + (7d)(ab+bc+ca) + 2(ab²+bc²+ca²) + 4(a²b+b²c+c²a) +9abc) =

x³ + (-d²+3e)x + (d³ - 3d� + (2d)(d�-2e) + (7d)(e) + 2(ab²+bc²+ca²) + 4(a²b+b²c+c²a) +9abc) =

 . . . . . . I'm not sure there is a solution to this...  The bold green expression, above, factors as (2a+b)(2b+c)(2c+a), which isn't helpful.

 


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