Question

If √2 is a zero of 6x³+√2x²-10x-4√2, find its other zeroes.

Answer 1

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.

HOPE THIS WILL HELP YOU....

Answer 2

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


Missing image in answer given at top of answer.
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