### 6 Answers

- sepiaLv 71 year agoFavourite answer
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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- ComoLv 71 year ago
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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- TomVLv 71 year ago
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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- Mike GLv 71 year ago
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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- VamanLv 71 year ago
x^3+4x^2+x-6=0= x^2(x+3) + (x^2+x-6)= x^2(x+3) + (x+3)(x-2)=(x+3) ( x^2+x-2)= (x+3) (x+2)(x-1)

The roots are -3,1,-2

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- PopeLv 71 year ago
Look for the little integers first. Clearly 1 is a solution. When the polynomial on the left is divided by (x - 1) the result must be a quadratic. Factor that, and you will have the given cubic polynomial completely factored. Use the zero-product law to find the three roots.

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