Math, asked by meghana5422, 7 months ago

if 1,-1 and 2 are zeroes of a polynomial p(x)=x3- ax2 -bx+2 find the values of a and b


plz answer this question as soon as possible this is important
who will give answer correctly I will follow them give thanks mark as brainliest and rate you

unsual answers will be reported
plz do it fast​

Answers

Answered by snehitha2
2

Answer :

a = 2, b = 1

Step-by-step explanation :

Can be solved in two ways,

  • substituting the values of zeroes in place of x
  • from the relationship between zeroes and coefficients

_______________________

Method - 1 :

    Given polynomial,

   p(x) = x³ - ax² - bx + 2

>>> 1,-1 and 2 are zeroes of a polynomial

So, when we substitute the above values in place of x ; the result is zero.

  • Put x = 1,

         p(1) = 0

 1³ - a(1)² - b(1) + 2 = 0

   1 - a - b + 2 = 0

       3 - a - b = 0

       \boxed{\bf a+b=3}

  • Put x = -1,

        p(-1) = 0

(-1)³ - a(-1)² - b(-1) + 2 = 0

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

 -1 - a + b + 2 = 0

     - a + b + 1 = 0

       \boxed{\bf b-a=-1}

  • Put x = 2,

        p(2) = 0

2³ - a(2)² - b(2) + 2 = 0

 8 - a(4) - 2b + 2 = 0

 8 - 4a - 2b + 2 = 0

    4a + 2b = 10

    2(2a+b) = 2(5)

     \boxed{\bf 2a+b=5}

______________________

We got three equations,

  a + b = 3  --(1)

  b - a = -1  --(2)

2a + b = 5 --(3)

Equation (1)

   a + b = 3

     a = 3 - b

substitute the value of a in equation (2)

Equation (2)

   b - a = -1

  b - (3-b) = -1

  b - 3 + b = -1

   2b = 3 - 1

   2b = 2

   b = 2/2

    b = 1

=> a = 3 - b

       = 3 - 1

       = 2

Substitute the values of a and b in equation (3) to VERIFY

 >> 2a + b

      2(2) + 1

      4 + 1

      5

Therefore, a = 2, b = 1

____________________

Method - 2 :

        For a cubic polynomial of the form,

             ax³ + bx² + cx + d  

Let α,β,γ are the zeroes

RELATIONSHIP BETWEEN ZEROES AND COEFFICIENTS :

➤  α + β + γ = -b/a

➤ αβ + βγ + αγ = c/a

➤ αβγ = -d/a

  Given polynomial,

   x³ - ax² - bx + 2

and its zeroes : 1,-1,2

From the above relation,

=> 1 + (-1) + 2 = -(-a)/1

    1 - 1 + 2 = a

        a = 2

=> (1)(-1) + (-1)(2) + (2)(1) = -b/1

     -1 - 2 + 2 = -b

        -b = -1

         b = 1

∴ a = 2, b = 1

Similar questions