Math, asked by ash14774, 6 months ago

find the zeros of the polynomial p(x)=x³-12x²+39x-28, if it is given that the zeroes are a-d, a and a+d​

Answers

Answered by snehitha2
0

Answer :

The zeroes are 1,4 and 7.

Given :

→ p(x) = x³-12x²+39x-28

→ the zeroes are a-d, a and a+d​

To find :

→ Zeroes of the polynomial

Solution :

We must be knowing, the relationship between zeroes and coefficients to find the zeroes.

For a polynomial of the form ax³+bx²+cx+d

The relationship between zeroes and coefficients of a cubic polynomial :

⇒ Sum of zeroes = -b/a

⇒ Product of zeroes = -d/a

-----------------------------------------------------

Given polynomial,

p(x) = x³-12x²+39x-28

║a = 1, b = -12, c = 39, d = -28

>> Sum of zeroes = -(-12)/1

       a-d + a + a+d = 12

             3a = 12

               a = 12/3

                a = 4

>> Product of zeroes = -(-28)/1

    (a-d)(a)(a+d) = 28

       (a² - d²)(a) = 28

        (4² - d²)(4) = 28

          16 - d² = 28/4

           16 - d² = 7

            d² = 16-7

            d² = 9

            d = √9

            d = 3

-------------------------------------------------

>>> The zeroes are a-d , a , a+d

      ⇒ a-d = 4-3 = 1

      ⇒ a = 4

      ⇒ a+d = 4+3 = 7

The zeroes of the polynomial p(x) = x³-12x²+39x-28 are 1,4 and 7

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