Question

Find a cubic polynomial whose zeros are 1/2, 1 and -3​

Answer 1

Given

• $\sf{Zeros ~of~ Cubic~ polynomial ~are1/2,3and-3.}$

To find

• Cubic Polynomial

Let,

• alpha=1/2
• beta=1
• gamma=-3

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$\\{\sf Sum \: of \:zeroes \: = \alpha + \beta + \gamma }$

${\sf Sum \: of \:zeroes \: = \frac{1}{2} + 1 - 3 }$

${\sf Sum \: of \:zeroes \: = \frac{1 + 2 - 3}{2} }$

${\sf Sum \: of \:zeroes \: = 0 }$

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$\\{\sf Sum \: of \: Product \: of \: zeros= \alpha \beta + \beta \gamma + \gamma \alpha }$

${\sf Sum \: of \: Product \: of \: zeros= \frac{1}{2} \times 1 + 1 \times (- 3 )+ (3) \times \frac{1}{2} }$

${\sf Sum \: of \: Product \: of \: zeros= \frac{1}{2} - 3 - \frac{3}{2} }$

${\sf Sum \: of \: Product \: of \: zeros= \frac{1 - 6 - 3}{2} }$

${\sf Sum \: of \: Product \: of \: zeros= \frac{ - 8}{2} }$

${\sf Sum \: of \: Product \: of \: zeros= - 4 }$

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$\\{\sf Product \: of \: zeros= \alpha \beta \gamma }$

${\sf Product \: of \: zeros= \frac{1}{2} \times 1 \times - 3 }$

${\sf Product \: of \: zeros= \frac{ - 3}{2} }$

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We know that:-

• A cubic polynomial when the sum of zeros , sum of products of zeros and Product of zeros are Given by

$\\\\{ \sf{f(x)=k \{ {x}^{3} - (sum \: of \: zeros) {x}^{2} + (sum \: of \: products \: of \: zeros)x - Product \: of \: zeros \}}}$

${ \sf{f(x)=k \{ {x}^{3} - (0) {x}^{2} + ( - 4)x - ( \frac{ - 3}{ \: \: 2} ) \}}}$

${ \sf{f(x)=k \{ {x}^{3} - 4x + \frac{ 3}{ 2}\}}}$

• The Required Cubic polynomial f(x)=x³-4x+3/2.