Question
Sat July 21, 2012 By:

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)

Expert Reply
Sun July 22, 2012
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 
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