Prove that -6, -frac{1}{2} and 1 are the zero | vedclass

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(A) Given polynomial: $p(x) = 2x^3 + 11x^2 - 7x - 6$. Step $1$: Check if $-6, -\frac{1}{2}, 1$ are zeros. $p(-6) = 2(-6)^3 + 11(-6)^2 - 7(-6) - 6 = 2(

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Zeros are $-6, -\frac{1}{2}, 1$; Relationship verified.

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(A) Given polynomial: $p(x) = 2x^3 + 11x^2 - 7x - 6$. Step $1$: Check if $-6, -\frac{1}{2}, 1$ are zeros. $p(-6) = 2(-6)^3 + 11(-6)^2 - 7(-6) - 6 = 2(-216) + 11(36) + 42 - 6 = -432 + 396 + 42 - 6 = 0$. $p(-\frac{1}{2}) = 2(-\frac{1}{8}) + 11(\frac{1}{4}) - 7(-\frac{1}{2}) - 6 = -\frac{1}{4} + \frac{11}{4} + \frac{7}{2} - 6 = \frac{10}{4} + \frac{14}{4} - 6 = 6 - 6 = 0$. $p(1) = 2(1)^3 + 11(1)^2 - 7(1) - 6 = 2 + 11 - 7 - 6 = 0$. Thus,$-6, -\frac{1}{2}, 1$ are zeros. Step $2$: Verify relationship. Sum of zeros: $\alpha + \beta + \gamma = -6 - \frac{1}{2} + 1 = -5 - 0.5 = -5.5 = -\frac{11}{2} = -\frac{	ext{coefficient of } x^2}{	ext{coefficient of } x^3}$. Sum of product of zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (-6)(-\frac{1}{2}) + (-\frac{1}{2})(1) + (1)(-6) = 3 - 0.5 - 6 = -3.5 = -\frac{7}{2} = \frac{	ext{coefficient of } x}{	ext{coefficient of } x^3}$. Product of zeros: $\alpha\beta\gamma = (-6)(-\frac{1}{2})(1) = 3 = -\frac{-6}{2} = -\frac{	ext{constant term}}{	ext{coefficient of } x^3}$.

Prove that $-6, -\frac{1}{2}$ and $1$ are the zeros of the cubic polynomial $p(x) = 2x^3 + 11x^2 - 7x - 6$. Also,verify the relationship between the zeros and the coefficients.

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