If √2 is zero of the cubic polynomial 6x3 + 2x2 - 10x - 42, the find it’s other two zeroes.


Let P(x) = 6x3 + 2x2 - 10x - 42


 


As √2 is one of the zeroes of P(x).


 


g(x) = (x - √2) is one of the factors of P(x).


 



 


By division algorithm


 


Dividend = (divisor) (quotient) + remainder


 


i.e. p(x) = g(x)q(x) + r(x)


 


clearly r(x) = 0 and q(x) = 6x2 + 7√2 + 4


 


6x3 + √2x2 - 10x - 4√2 = (x - √2) (6x2 + 7√2 + 4)


 


6x3 + √2x2 - 10x - 4√2 = 0


 


= (x - √2) (6x2 + 7√2 + 4) = 0


 


= (x - √2) {6x2 + (3√2x + 4√2x) + 4} = 0 (by splitting the middle term)


 


= (x - √2) {6x2 + 3√2x + 4√2x + 4} = 0


 


= (x - √2) {3√2x (√2x + 1) + 4(√2x + 1)} = 0


 


= (x - √2) (√2x + 1) (3√2x + 4) = 0


 


x = - 1/√2, 2√2/3, √2


 

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