# Find the sum of 3 roots of eqn x^3+4x^2+x-6=0?

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x^3 + 4x^2 + x - 6= 0

(x - 1) (x + 2) (x + 3) = 0

The sum of 3 roots:

-3 - 2 + 1 = -4

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• f (1) = 1 + 4 + 1 - 6 = = 0

Thus x - 1 is a factor.

Find other factors by using synthetic division :--

:-

1 | 1_________4________1_______- 6

__| __________1________5________6

__| 1_________5________6________0

( x - 1 ) (x² + 5x +. 6) = 0

( x - 1 ) ( x + 3 ) ( x + 2 ) = 0

x = 1 , - 3 , - 2

Sum of roots = - 4

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• It's not necessary to find the individual roots:

(x-a)(x-b)(x-c) = 0

(x-a)(x²-bx-cx+bc) = 0

x³ -ax² - bx² - cx² + bcx+abx+acx - abc = 0

x° - (a+b+c)x² + (bc+ab+ac)x - abc = 0

x³ + 4x² + x - 6 = 0

Comparing coefficients of like terms we see:

-(a+b+c) = 4

(bc+ab+ac) = 1

-abc = -6

Ans:

From which we see that the sum of the roots is:

(a+b+c) = -4

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• x = 1 is a root by inspection so (x-1) is a factor

Division by (x-1) gives (x^+5x+6)

(x-1)(x+3)(x+2) = 0

x = -3, - 2, 1

Sum = -4

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