Question

If it is given that root3 is a zero of a polynomial 6x³+root2x²-10x-4root2 find the other zeroes.

Answer 1

Step-by-step explanation:

6x^3+root2x^2-10x-4 ------------------(1)

x=2

put the value in eqn (1),

6(2)^3+√2(2)^2-10(2)-4

6(8)+4√2-20-4

48+√2-24

24+√2

Answer 2

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

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

Divide p(x) by x-√2

x-√2) 6x³+√2x²-10x-4√2 (6x² +7√2x

6x³-6√2x²

(-) (+)

----------------------------

7√2x² -10x-4√2

7√2x² -14x

(-) (+)

-------------------------

4x - 4√2

4x - 4√2

(-) (+)

---------------------

0

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

= (x-√2) (6x² +4√2x + 3√2x + 4)

[By splitting middle term]

= (x-√2) [ 2x(3x+2√2) + √2(3x+2√2)

= (x-√2) (2x+√2) (3x+2√2)

For zeroes of p(x), put p(x)= 0

(x-√2) (2x+√2) (3x+2√2)= 0

x= √2 , x= -√2/2 ,x= -2√2/3

x= √2 , x= -1 /√2 ,x= -2√2/3

[ Rationalising second zero]

Hence, the other two zeroes of p(x) are -1/√2 and -2√2/3.

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