Roots of Polynomial

Algebra Level 3

If a , b , c a,b,c are the distinct roots of x 3 − 7 x 2 − 6 x + 5 = 0 x^3-7x^2-6x+5=0 ,

Find the value of ( a + b ) ( b + c ) ( a + c ) (a+b)(b+c)(a+c) .


The answer is -37.

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Ikkyu San by Ikkyu San , 23, Thailand

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3 solutions

f ( x ) = x 3 − 7 x 2 − 6 x + 5 = ( x − a ) ( x − b ) ( x − c ) f(x) = x^3 - 7x^2 - 6x + 5 = (x -a )(x-b)(x-c)

Sum of roots is 7 7 .

7 = a + b + c ⇒ a + b = 7 − c , b + c = 7 − a , a + c = 7 − b 7 = a +b + c \Rightarrow a + b = 7 -c , b+c = 7-a , a + c = 7 -b

⇒ ( a + b ) ( b + c ) ( a + c ) = ( 7 − a ) ( 7 − b ) ( 7 − c ) = f ( 7 ) = − 37 \Rightarrow (a+b)(b+c)(a+c) = (7-a)(7-b)(7-c) = f(7) = \boxed{-37}

10 Helpful 0 Interesting 1 Brilliant 0 Confused

Just use the fact that - (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc

Aditya Kumar - 5 years ago
Benjamin Ononogbu
Sep 25, 2015

using vieta's formular (a+b)(b+c)(a+c)=(a+b+c)(ab+bc+ac)-(abc) having known that a+b+c=7, ab+bc+ac=-6 and abc=-5. substitute in the formular to get -37

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Aditya Chauhan
Aug 22, 2015

a + b + c = 7 a+b+c = 7

a b + b c + c a = − 6 ab+bc+ca = -6

a b c = − 5 abc =-5

\((a+b)(b+c)(c+a)

= (7-c)(7-a)(7-b)\)

= 343 − 49 ( a + b + c ) + 7 ( a b + b c + c a ) − a b c =343-49(a+b+c) +7(ab+bc+ca)-abc

= 343 − 49 ( 7 ) + 7 ( − 6 ) − ( − 5 ) =343-49(7)+7(-6)-(-5) = − 37 \boxed{-37}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You almost reached my solution. Check mine. It is a bit different at the end.

Vishwak Srinivasan - 5 years, 9 months ago

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