Question

if zeros of polynomial x 3 + 15 x 2 + 66 x + 80 are in AP and write zeroes as first three terms of AP and find the 10th term​

Answer 1

$\underline\pink{\frak{given}}$

polynomial given is

${x}^{3} + 15 {x}^{2} + 66x + 80$

zeroes of this polynomial are in AP

$\underline{\pink{\frak{to \: find}}}$

$tenth \: term \: of \: \\ the \: arithematic \: progression \\ a_{10}$

$\underline{\pink{\frak{solution}}}$

As we can see that,

degree of given polynomial is 3

hence it is a cubic polynomial having 3 zeroes.

In the given polynomial

a = 1 ; b = 15 ; c = 66 ; d = 80

Let,

zeroes of polynomial be

m-1 ; m ; m+1

so,

in a cubic polynomial

sum of zeroes = -b / a

so,

m-1+ m +m +1 = -15 / 1

3 m = -15

$\boxed {m = - 5}$

so,

the zeores will be,

m -1 = -5-1 = - 6

m = - 5

m + 1 = -5+1 = - 4

AP: -6 , -5 , -4

so,

first term of AP, a = -6

common difference, d = 1

so,

$a_{10 }= a + (n - 1)d \\ \\ a_{10} = - 6 + 9(1) \\ \\ \boxed {a_{10 }= 3}$

Answer 2

Step-by-step explanation:

polynomial given is

{x}^{3} + 15 {x}^{2} + 66x + 80x

3

+15x

2

+66x+80

zeroes of this polynomial are in AP

\underline{\pink{\frak{to \: find}}}

tofind

\begin{gathered}tenth \: term \: of \: \\ the \: arithematic \: progression \\ a_{10}\end{gathered}

tenthtermof

thearithematicprogression

a

10

\underline{\pink{\frak{solution}}}

solution

As we can see that,

degree of given polynomial is 3

hence it is a cubic polynomial having 3 zeroes.

In the given polynomial

a = 1 ; b = 15 ; c = 66 ; d = 80

Let,

zeroes of polynomial be

m-1 ; m ; m+1

so,

in a cubic polynomial

sum of zeroes = -b / a

so,

m-1+ m +m +1 = -15 / 1

3 m = -15

\boxed {m = - 5}

m=−5

so,

the zeores will be,

m -1 = -5-1 = - 6

m = - 5

m + 1 = -5+1 = - 4

AP: -6 , -5 , -4

so,

first term of AP, a = -6

common difference, d = 1

so,

\begin{gathered}a_{10 }= a + (n - 1)d \\ \\ a_{10} = - 6 + 9(1) \\ \\ \boxed {a_{10 }= 3}\end{gathered}

a

10

=a+(n−1)d

a

10

=−6+9(1)

a

10

=3