Question
Find a cubic polynomial whose zeros are $3, 5$ and $-2$
✓
Answer
Let $\alpha,\ \beta$ and $\gamma$ be the zeros of the required polynomial.
Then we have:
$\alpha+\beta+\gamma$
$=3+5+(-2)=6$
$\alpha\beta+\beta\gamma+\gamma\alpha$
$=3\times5+5\times(-2)+(-2)\times3=-1$
and $\alpha\beta\gamma=3\times5\times-2=-30$
Now, $\text{p}(\text{x})=\text{x}^3-\text{x}^2(\alpha+\beta+\gamma)+\text{x}(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma$
$=\text{x}^3-\text{x}^2\times6+\text{x}\times(-1)-(- 30)$
$=\text{x}^3-\text{6x}^2-\text{x}=30$
So, the required polynomial is $p(x)=x^3-6 x^2-x+30$
Then we have:
$\alpha+\beta+\gamma$
$=3+5+(-2)=6$
$\alpha\beta+\beta\gamma+\gamma\alpha$
$=3\times5+5\times(-2)+(-2)\times3=-1$
and $\alpha\beta\gamma=3\times5\times-2=-30$
Now, $\text{p}(\text{x})=\text{x}^3-\text{x}^2(\alpha+\beta+\gamma)+\text{x}(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma$
$=\text{x}^3-\text{x}^2\times6+\text{x}\times(-1)-(- 30)$
$=\text{x}^3-\text{6x}^2-\text{x}=30$
So, the required polynomial is $p(x)=x^3-6 x^2-x+30$
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