Question

The product of the zeroes of x^3+4x^2+x-6 is

Answer 1

$Solution$

${x}^{3} + 4 {x}^{2} + x - 6 \\ \\ = > product \: of \: zeroes = \frac{ - d}{a} \: \\ = > \alpha . \beta .\gamma = \frac{ - d}{a} \\ \\ = > \alpha . \beta .\gamma = \frac{ - ( - 6)}{1} \\ \\ = > \alpha . \beta .\gamma = 6$

$Hope \:it \:helps$

Answer 2

Answer:

We know that,

$\alpha \beta \gamma = - \frac{d}{a}$

Now,

$\alpha \beta \gamma = - \frac{ - 6}{1}$

$\alpha \beta \gamma = 6$

Therefore, the product of zeroes is 6.

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