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

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  • sepia
    Lv 7
    1 year ago
    Favourite 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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  • Como
    Lv 7
    1 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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  • TomV
    Lv 7
    1 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 G
    Lv 7
    1 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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  • Vaman
    Lv 7
    1 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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  • Pope
    Lv 7
    1 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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