CBSE Class 10 Answered

if underroot 2 and -underroot 2 are the zeroes of p(x) = 2x^$ + 7x^3 - 19x^2 - 14x + 30, then find the other zeroes of p(x)

Asked by | 21 Jul, 2012, 06:55: PM

Expert Answer

Answer : Given : If underroot 2 and -underroot 2 are the zeroes of p(x) = 2x⁴ + 7x³ - 19x² - 14x + 30

To find : the other zeroes of p(x)

Since we know that -underrroot 2 is a zero of p(x) , p(-underrroot 2) = 0

=> x- (-underrroot 2 ) is a factor of p(x)

dividing 2x⁴ + 7x³ - 19x² - 14x + 30 by x-(-underrroot 2) i.e x+underrroot 2, we get

we get the quotient = 2x³ + (7 - (2 * ( underroot 2 ) ) ) x² - (15 + (7*underroot 2))x + 15 * underrroot 2

=> 2x⁴ + 7x³ - 19x² - 14x + 30 = (x+underrroot 2)

* (2x³ + (7 - (2 * ( underroot 2 ) ) ) x² - (15 + (7*underroot 2))x + 15 * underrroot 2 )

Also underroot 2 is a zero of of 2x³ + (7 - (2 * ( underroot 2 ) ) ) x² - (15 + (7*underroot 2))x + 15 * underrroot 2

therefore underroot 2 is the factor of 2x³ + (7 - (2 * ( underroot 2 ) ) ) x² - (15 + (7*underroot 2))x + 15 * underrroot 2 =0

By dividing 2x³ + (7 - (2 * ( underroot 2 ) ) ) x² - (15 + (7*underroot 2))x + 15 * underrroot 2 by (x - underroot 2 ) , the quotient is 2x² + 7x - 15

Therefore p(x) i.e 2x⁴ + 7x³ - 19x² - 14x + 30 = (x + underoot 2 ) * (x -underoot 2 ) * ( 2x² + 7x - 15 )

and 2x² + 7x - 15 = ( 2x - 3 ) ( x + 5)

=> (2x⁴ + 7x³ - 19x² - 14x + 30 ) = (x + underoot 2 ) * (x -underoot 2 ) * ( 2x - 3 ) * ( x+5 )

=> All the zeroes of p(x) are

underoot 2 , -underoot 2 , 3/2 , -5

=> The other roots are 3/2 and -5 Answer

Answered by | 22 Jul, 2012, 12:40: AM

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