Polynomial Roots

Algebra Level 4

Suppose that a + b , b + c , a+b,b+c, and c + a c+a are the complex roots of the equation x 3 + 10 x 2 − 4 x + 23 = 0. x^3+10x^2-4x+23=0. If the 3rd degree polynomial that has a , b a,b and c c as roots can be written in the form x 3 + s x 2 + t x + u , x^3+sx^2+tx+u, what is the value of s + t + u s+t+u ?


The answer is -192.

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Xiaoma Zhao by Xiaoma Zhao , 16, USA

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2 solutions

Steven Yuan
May 21, 2018

We have ( a + b ) + ( b + c ) + ( c + a ) = − 10 (a + b) + (b + c) + (c + a) = -10 by Vieta's, so a + b + c = − 5. a + b + c = -5. Thus, the roots of the original polynomial can also be expressed as − a − 5 , − b − 5 , − c − 5. -a - 5, -b - 5, -c - 5.

Let f ( x ) = x 3 + 10 x 2 − 4 x + 23. f(x) = x^3 + 10x^2 - 4x + 23. The polynomial with roots a + 5 , b + 5 , c + 5 , a + 5, b + 5, c + 5, which are opposites of the roots of f ( x ) , f(x), is

g ( x ) = − f ( − x ) = − [ ( − x ) 3 + 10 ( − x ) 2 − 4 ( − x ) + 23 ] = x 3 − 10 x 2 − 4 x − 23 g(x) = -f(-x) = -[(-x)^3 + 10(-x)^2 - 4(-x) + 23] = x^3 - 10x^2 - 4x - 23

(we take the opposite of the intended f ( − x ) f(-x) since our final answer must have a positive leading coefficient). Next, the polynomial with roots a , b , c , a, b, c, which are 5 less than the roots of g ( x ) , g(x), is

h ( x ) = g ( x + 5 ) = ( x + 5 ) 3 − 10 ( x + 5 ) 2 − 4 ( x + 5 ) − 23. h(x) = g(x + 5) = (x + 5)^3 - 10(x + 5)^2 - 4(x + 5) - 23.

Instead of expanding this out completely, we use a shortcut: use synthetic division on the coefficients of g ( x ) g(x) repeatedly, removing the final coefficient calculated each time, until there are no more coefficients to divide. The removed coefficients are the coefficients of h ( x ) . h(x). The method looks like this:

5 1 10 − 4 23 − 10 0 40 5 1 − 5 − 29 − 168 5 0 5 1 0 − 29 5 5 1 5 1 \begin{array}{c|cccc} 5 & 1 & 10 & -4 & 23 \\ & & -10 & 0 & 40 \\ \hline 5 & 1 & -5 & -29 & \boxed{-168} \\ & & 5 & 0 & \\ \hline 5 & 1 & 0 & \boxed{-29} & \\ & & 5 & & \\ \hline 5 & 1 & \boxed{5} & & \\ \hline & \boxed{1} \end{array}

Thus, h ( x ) = x 3 + 5 x 2 − 29 x − 168 , h(x) = x^3 + 5x^2 - 29x - 168, and s + t + u = 5 − 29 − 168 = − 192 . s + t + u = 5 - 29 - 168 = \boxed{-192}.

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Nice solution although I didn't understand how your synthetic division shortcut works.

Xiaoma Zhao - 3 years ago
Xiaoma Zhao
May 20, 2018

We need to find the values of

  • s = − ( a + b + c ) s=-(a+b+c)
  • t = a b + b c + a c t=ab+bc+ac
  • u = − a b c . u=-abc.

We can use Vieta's formulas to get

  1. a + b + b + c + c + a = − 10 a+b+b+c+c+a=-10
  2. ( a + b ) ( b + c ) + ( b + c ) ( c + a ) + ( c + a ) ( a + b ) = − 4 (a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)=-4
  3. ( a + b ) ( b + c ) ( c + a ) = − 23. (a+b)(b+c)(c+a)=-23.

From the first equation, we get 2 ( a + b + c ) = − 10 ⟹ a + b + c = − 5. 2(a+b+c)=-10\implies a+b+c=-5. Therefore, we have s = − ( a + b + c ) = 5. s=-(a+b+c)=5. We can also write the following equations

  • a + b = − 5 − c a+b=-5-c
  • b + c = − 5 − a b+c=-5-a
  • c + a = − 5 − b . c+a=-5-b.

Now, we can plug these in the second and third equations to get

  • ( − 5 − c ) ( − 5 − a ) + ( − 5 − a ) ( − 5 − b ) + ( − 5 − b ) ( − 5 − c ) = − 4 (-5-c)(-5-a)+(-5-a)(-5-b)+(-5-b)(-5-c)=-4
  • ( − 5 − c ) ( − 5 − a ) ( − 5 − b ) = − 23. (-5-c)(-5-a)(-5-b)=-23.

We can simplify and rearrange both equations to get

  • a b + b c + c a = − 10 ( a + b + c ) − 79 ab+bc+ca=-10(a+b+c)-79
  • a b c + 5 ( a b + b c + a c ) = − 25 ( a + b + c ) − 102 abc+5(ab+bc+ac)=-25(a+b+c)-102

Since we know that a + b + c = − 5 , a+b+c=-5, we know that

  • a b + b c + c a = − 10 ⋅ ( − 5 ) − 79 = − 29 ab+bc+ca=-10\cdot (-5)-79=-29
  • a b c + 5 ( a b + b c + a c ) = − 25 ⋅ ( − 5 ) − 102 = 23 abc+5(ab+bc+ac)=-25\cdot (-5)-102=23

Therefore, t = a b + b c + a c = − 29. t=ab+bc+ac=-29. We can plug this into the second equation to get − u = a b c = 168 ⟹ u = − 168. -u=abc=168\implies u=-168.

Finally, we have s + t + u = 5 − 29 − 168 = − 192 . s+t+u=5-29-168=\boxed{-192}.

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