हिंदी

Given that 2 is a zero of the cubic polynomial 6x3+2x2-10x-42, find its other two zeroes.

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प्रश्न

Given that `sqrt(2)` is a zero of the cubic polynomial `6x^3 + sqrt(2)x^2 - 10x - 4sqrt(2)`, find its other two zeroes.

योग
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उत्तर

Given, `sqrt(2)` is one of the zero of the cubic polynomial.

Then, `(x - sqrt(2))` is one of the factor of the given polynomial p(x) = `6x^3 + sqrt(2)x^2 - 10x - 4sqrt(2)`.

So, by dividing p(x) by `x - sqrt(2)`

                   `6x^2 + 7sqrt(2)x + 4`
`(x - sqrt(2))")"overline(6x^3 + sqrt(2)x^2 - 10x - 4sqrt(2))`
                    `6x^3 - 6sqrt(2)x^2`
                     –     +                                   
                               `7sqrt(2)x^2 - 10x - 4sqrt(2)`
                               `7sqrt(2)x^2 - 14x`
                               –      +                 
                                        `4x - 4sqrt(2)`
                                        `4x - 4sqrt(2)`   
                                                           
                                               0

`6x^3 + sqrt(2)x^2 - 10x - 4sqrt(2) = (x - sqrt(2)) (6x^2 + 7sqrt(2)x + 4)`

By splitting the middle term,

We get,

`(x - sqrt(2)) (6x^2 + 4sqrt(2)x + 3sqrt(2)x + 4)`

= `(x - sqrt(2)) [2x(3x + 2sqrt(2)) + sqrt(2)(3x + 2sqrt(2))]`

= `(x - sqrt(2)) (2x + sqrt(2)) (3x + 2sqrt(2))`

To get the zeroes of p(x),

Substitute p(x) = 0

`(x - sqrt(2)) (2x + sqrt(2)) (3x + 2sqrt(2))` = 0

`x = sqrt(2) , x = -sqrt(2)/2, x = (-2sqrt(2))/3`

Hence, the other two zeroes of p(x) are `-sqrt(2)/2` and `(-2sqrt(2))/3`.

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अध्याय 2: Polynomials - Exercise 2.4 [पृष्ठ १५]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 10
अध्याय 2 Polynomials
Exercise 2.4 | Q 3 | पृष्ठ १५

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