Question

Given that the zeroes of the cubic polynomial x³ – 6x² + 3x +10 are of the form a,a + b,a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.


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Answer 1

GIVEN :–

• A cubic polynomial x³ – 6x² + 3x + 10 = 0 have three roots a , a+b , a+2b.

TO FIND :–

• Value of a , b = ?

SOLUTION :–

• Let's use Hit & trail method –

• By putting x = -1 cubic polynomial gives '0' . Hence , x + 1 is a factor of cubic polynomial.

⇒ (x + 1)(x² - 7x + 10) = 0

⇒ (x + 1)(x² - 5x - 2x + 10) = 0

⇒ (x + 1)[x(x - 5) - 2(x - 5)] = 0

⇒ (x + 1)[(x - 2)(x - 5)] = 0

⇒ (x + 1)(x - 2)(x - 5) = 0

⇒ x = -1 , 2 , 5

• Compare with given roots –

⇒ a = -1 , a + b = 2 & a + 2b = 5

• So that –

⇒ a + b = 2

⇒ (-1) + b = 2

⇒ b = 2 + 1

⇒ b = 3

☆ Hence , The value of a & b is - 1 & 3 .

Answer 2

$\star\bold\pink{ \: Given \: cubic \: polynomial \: is }$

p (x) = x³ − 6x² + 3x + 10

The zeros of the polynomial p (x) are of the form a, a + b and a + 2b 

Then,

$\bold{a+a+b+a+2b= \frac{ - 6}{1} }$

=>3a+3b=6

=>a+b=2     ----------------(i)

$\bold{ Also,  a(a+b)+(a+b)(a+2b)+a(a+2b)= \frac{3}{1} }$

=>a² + ab + a² + 2ab + ab + 2b² + a² + 2ab = 3

=> 3a² + 2b² + 6ab = 3  ----(ii)

$\bold{and   a(a+b)(a+2b) = - \frac{ - 10}{1} }$

=> a³ + a2b + 2a²b +2ab² = 10

From (i), b = 2 − a

Putting this value in (ii), we get

=> 3a² + 2 ( 2 − a )² + 6a ( 2 − a ) = 3

=> 3a² + 2 ( 4 − 4a + a² ) + 12a − a² = 3

=> −a² + 4a+5=0

=> a² − 4a−5=0

=> a2−5a+a−5=0

=>a(a−5)+(a−5)=0

=> .(a−5)(a+1)=0

=> a=5  or   a=−1

=> b=−3 or   b=3 respectively

From equation (iii), we get

at a=5, b=−3

=> 53+(5)²(−3)+2(5)2(−3)+²(5)(−3)²=−10

=> 125−75−150+90=−10

=> −10=−10 which is true.

and at a =−1, b=3, we have

=> ( − 1 ) ³ + ( −1 ) ² 3 + 2 ( −1 ) ² ( 3 ) + 2 ( −1 ) (3)² =−10

=> −1+3−6−18=10

=> − 22 = −10  which is not true.

Thus, a = 5, b = − 3

Zeros of the polynomial are 5, 5 − 3, 5 − 2 × 3 ie 5 , 2 , −1

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