Question
Find a cubic polynomial whose zeroes are $\frac{1}{2},$ $1$ and $−3$.
✓
Answer
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as:
$x^3- (a + b + c)x^2 + (ab + bc + ca)x - abc ...(1)$
Let $\text{a}=\frac{1}{2},$ $b = 1$ and $c = -3$
Substituting the values in (1), we get
$\text{x}^3-\Big(\frac{1}{2}+1-3\Big)\text{x}^2+\Big(\frac{1}{2}-3-\frac{3}{2}\Big)\text{x}-\Big(\frac{-3}{2}\Big)$
$\Rightarrow\text{x}^3-\Big(\frac{-3}{2}\Big)\text{x}^2-\text{4x}+\frac{3}{2}$
$\Rightarrow\text{2x}^3+\text{3x}^2-\text{8x}+3$
$x^3- (a + b + c)x^2 + (ab + bc + ca)x - abc ...(1)$
Let $\text{a}=\frac{1}{2},$ $b = 1$ and $c = -3$
Substituting the values in (1), we get
$\text{x}^3-\Big(\frac{1}{2}+1-3\Big)\text{x}^2+\Big(\frac{1}{2}-3-\frac{3}{2}\Big)\text{x}-\Big(\frac{-3}{2}\Big)$
$\Rightarrow\text{x}^3-\Big(\frac{-3}{2}\Big)\text{x}^2-\text{4x}+\frac{3}{2}$
$\Rightarrow\text{2x}^3+\text{3x}^2-\text{8x}+3$
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