Question

if two zeros of the polynomial x to the power 4 minus 6 x cube minus 26 X square + 138 x minus 35 are 2 plus root 3 find the other zeroes

Answer 1

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Solution ≈≈
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Let f (x) = x^4 - 6x^3 - 26x^2 + 138x - 35

$let \: \: \alpha = (2 + \sqrt{3} ) \: \: and \: \beta = (2 - \sqrt{3} ) \\ then \\ ( \alpha + \beta ) = 4 \: \: and \: \: \alpha \beta = (4 - 3) = 1.$
so, the quadratic Polynomial whose roots are Alpha and beta is given by

${x}^{2} - ( \alpha + \beta )x \: + \: \alpha \beta = ( {x}^{2} - 4x + 1).$
=> (x^2 - 4x + 1 ) is a factor of f(x)

on dividing f (x) by (x^2 - 4x + 1 ), we get

Solution in above attachment⬆⬆

=> f (x) = (x^2 - 4x + 1 ) (x^2 - 2x - 35).

the other two zeroes of f (x) are given by (x^2 - 2x - 35) = 0

Now ,
${x}^{2} - 2x - 35 = 0 \\ = > {x}^{2} - 7x + 5x - 35 = 0 \\ = > x(x - 7) + 5(x - 7) = \\ = > (x - 7)(x + 5) = 0 \\ = > x = 7 \: \: or \: x = - 5$
Hence , the other two zeroes of f (x) are 7 and -5 .
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Answer 2

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