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Answer 1

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Answer 2

Step-by-step explanation:

1

:

Given: x³ - 6x² + 3x + 10

  Zeroes are of the form = a , a + b  , a + 2b

Value of a & b and all zeroes of the polynomial.

Sum of the roots = -coefficient of x²/ coefficient of x³

⇒ a + 2b + a + a + b = -(-6)/1                      

⇒ 3a + 3b = 6

⇒ 3 ( a + b ) = 6

⇒ a + b = 2 ..................(1)

Product of roots = -constant/coefficient of x³

⇒ ( a + 2b )( a + b ) a = -10/1

⇒ ( a + b + b )( a + b ) a = -10

⇒ ( 2 + b )( 2 ) a = -10   ( From eqaution (1) )

⇒ ( 2 + b ) 2a = -10

⇒ ( 2 + 2 - a ) 2a = -10

⇒ ( 4 - a ) 2a = -10

⇒ 4a - a² = -5

⇒ a² - 4a - 5 = 0

Now solving obtained quadratic equation we get,

a = 5 , -1                 

Now, Putting value of a in equation in Eqn (1)

We get,

When

a = 5   ⇒  5 + b = 2 ⇒ b = -3

a = -1   ⇒  -1 + b = 2 ⇒ b = 3

when a = 5 and b = -3,  Zeroes are 5 , ( 5 + (-3) ) = 2 , ( 5 + 2(-3) ) = -1

when a = -1 and b = 3 , Zeroes are -1 , ( -1 + 3 ) = 2 , ( -1 + 2(3) ) = 5

Therefore, Value of a = 5 & b = -3 or a = -1 & b = 3  and Zeroes are -1 , 2 , 5.

2.

√2 is zero of 6x³ +√2x² -10x -4√2 ,

So, (x - √2) is a factor of 6x³ + √2x² -10x -4√2 .

6x³ + √2x² -10x -4√2

= 6x³ -6√2x² + 7√2x² -14x + 4x - 4√2

= 6x²(x - √2) + 7√2x(x -2) + 4(x -√2)

= {6x² + 7√2x + 4}(x - √2)

= {6x² + 3√2x + 4√2x + 4}(x -√2)

= {3√2x(√2x + 1) + 4(√2x +1)}(x -√2)

= (3√2x +4)(√2x +1)(x - √2)

Hence, two other zeros are -4/3√2, and -1/√2