Question

find the all the zeros of the x^3+3x^2-2x-6 if two of its zeros are +√2​

Answer 1

${x}^{3} + 3 {x}^{2} - 2x - 6 = 0$

${x}^{2} (x + 3) - 2(x + 3) = 0$

$(x + 3)( {x}^{2} - 2) = 0$

$\implies x _{1} = 3$

$\implies x_{2} = \sqrt{2}$

$\implies x_{3} = - \sqrt{2}$

Answer 2

If two zeroes are √2 and -√2 so, the factor will be

(x-√2)(x+√2)=0

⇒x²-2=0

now, dividing the expression with this,

x²-2)x³+3x²-2x-6(x+3

x³-2x

- +

+3x²-6

+3x²-6

- +

x

so, x³+3x²-2x-6 = (x²-2) (x+3)

All the zeroes are = -√2 , +√2 , -3

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