If p, q and r are the zeroes of the cubic polynomial x^3 - 6x^2 - x + 30 then the product of the zeroes i .e. p.q.r. is
Answers
Step-by-step explanation:
p, q and r are the zeroes of the cubic polynomial x^3 - 6x^2 - x + 30 then the product of the zeroes i .e. p.q.r. is
Value of a = 5 & b = -3 or a = -1 & b = 3 and Zeroes are -1 , 2 , 5.
Step-by-step explanation:
Given: x³ - 6x² + 3x + 10
Zeroes are of the form = a , a + b , a + 2b
To find: Value of a & b and all zeroes of the polynomial.
Using relationship of coefficient and Zeroes of the polynomial we get,
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.
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