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Question -

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



Answer -

Given, √2 is one ofthe zero of the cubic polynomial.

Then, (x-√2) is one ofthe factor of the given polynomial p(x) = 6x³+√2x²-10x- 4√2.

So, by dividing p(x)by x-√2


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

By splitting themiddle term,

We get,

(x-√2) (6x² + 4√2x +3√2x + 4)

= (x-√2) [ 2x(3x+2√2)+ √2(3x+2√2)]

= (x-√2) (2x+√2)  (3x+2√2)

To get the zeroes ofp(x),

Substitute p(x)= 0

(x-√2) (2x+√2) (3x+2√2)= 0

x= √2 , x= -√2/2 ,x=-2√2/3

Hence, the other twozeroes of p(x) are -√2/2 and -2√2/3

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