InterviewSolution
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InterviewSolution
| 1. |
Find a cubic polynomial whose zeroes are 3,1/2and-1 |
| Answer» Let\xa0{tex}\\alpha{/tex} = 3, {tex}\\beta{/tex}\xa0=\xa0{tex}\\frac{1}{2}{/tex} and {tex}\\gamma{/tex}\xa0= -1. Then,{tex}( \\alpha + \\beta + \\gamma ) = \\left( 3 + \\frac { 1 } { 2 } - 1 \\right) = \\frac { 5 } { 2 }{/tex},{tex}( \\alpha \\beta + \\beta \\gamma + \\gamma \\alpha ) = \\left( \\frac { 3 } { 2 } - \\frac { 1 } { 2 } - 3 \\right) = \\frac { - 4 } { 2 }{/tex}\xa0= -2and\xa0{tex}\\alpha \\beta y = \\left\\{ 3 \\times \\frac { 1 } { 2 } \\times ( - 1 ) \\right\\} = \\frac { - 3 } { 2 }{/tex}The polynomial with zeros α,β and\xa0{tex}\\gamma{/tex} is:{tex}\\mathrm x^3-(\\mathrm\\alpha+\\mathrm\\beta+\\mathrm\\gamma)\\mathrm x^2+(\\mathrm{αβ}+\\mathrm{βγ}+\\mathrm{γα})\\mathrm x-\\mathrm{αβγ}{/tex}{tex}=\\mathrm x^3-\\frac52\\mathrm x^2-2\\mathrm x+\\frac32{/tex}Thus,\xa02x3- 5x2- 4x + 3 is the desired polynomial. | |