Question

If 1 and -1 are two zeroes of the Polynomial X3+2x2+ax+b,then find the values of a and b​

Answer 1

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GiveN:

• 1 and -1 are two zeroes of the Polynomial x³ + 2x² + ax + b

To FinD:

• Values of a and b?

Step-by-Step Explanation:

-1 is a zero of the p(x). Then, p(-1) = 0

Replacing x with -1:

⇒ x³ + 2x² + ax + b = 0 (when x = -1)

⇒ (-1)³ + 2(-1)² + a(-1) + b = 0

⇒ -1 + 2 - a + b = 0

⇒ 1 - a + b = 0

⇒ a - b = 1 ------------(1)

1 is also a zero of p(x). Then p(1) = 0

Replacing x with 1:

⇒ x³ + 2x² + ax + b = 0

⇒ (1)³ + 2(1)² + a(1) + b = 0

⇒ 3 + a + b = 0

⇒ a + b = -3 --------(2)

Adding (1) and (2),

⇒ a - b + a + b = 1 - 3

⇒ 2a = -2

⇒ a = -1

Then,

⇒ -1 - b = 1

⇒ b = -2

Hence,

The required value of a and b is:

$\Large{ \boxed{ \sf{ \red{a = - 1 \: and \: b = - 2}}}}$

Answer 2

Answer:

Given :

• If 1 and -1 are two zeroes of the Polynomial X3+2x2+ax+b

To Find :

• find the values of a and b

Solution :

If solving an equation, put it in standard form with 0 on one side and simplify.   [ details ]

• Know how many roots to expect.   [ details ]

• If you’re down to a linear or quadratic equation (degree 1 or 2), solve by inspection or the quadratic formula.   [ details ]

Then go to step 7.

• Find one rational factor or root. This is the hard part, but there are lots of techniques to help you.   [ details ]

• If you can find a factor or root, continue with step 5 below; if you can’t, go to step 6.

• Divide by your factor. This leaves you with a new reduced polynomial whose degree is 1 less.   [ details ]  

• For the rest of the problem, you’ll work with the reduced polynomial and not the original. Continue at step 3.

• If you can’t find a factor or root, turn to numerical methods.   [ details ]

Then go to step 7.

• If this was an equation to solve, write down the roots. If it was a polynomial to factor, write it in factored form, including any constant factors you took out in step 1.

______________________________

• Put both values of x in the polynomial.

If 1 are two zeroes of the Polynomial :

P( 1 ) = 1³ + 2( 1 )² + a ( 1 ) + b = 0

P( 1 ) = 1 + 2 ( 1 ) + a + b = 0

P( 1 ) = 1 + 2 + a + b = 0

P( 1 ) = 3 + a + b = 0

P( 1 ) = a + b = - 3

• P( 1 ) = a = - 3 - b ...... Equ ( 1 )

If -1 are two zeroes of the Polynomial :

P( - 1 ) = ( - 1 )³ + 2 ( - 1 )² + a ( - 1 ) + b = 0

P( - 1 ) = - 1 + 2 ( 1 ) - a + b = 0

P( - 1 ) = - 1 + 2 - a + b = 0

P( - 1 ) = 1 - a + b = 0

P( - 1 ) = - a + b = - 1

• P( - 1 ) = b = 1 + a ...... Equ ( 2 )

Substitute the value of b in equation 1

a = - 3 - ( 1 + a )

a= - 3 - 1 - a

a = - 4 - a

a + a = - 4

2a = - 4

a = - 4 / 2

a = - 2

substitute the value of a in equation 2 :

b = 1 + ( - 2 )

b = 1 - 2

b = - 1

• Hence the value of a is -2 and value of b is -1