Question

if the zeroes of the polynomial x^3+15x^2+66x+80 are in A.P.Find the 10th term of that particular A.P.​

Answer 1

GIVEN :

The zeroes of the polynomial $x^3+15x^2+66x+80$ are in A.P

TO FIND :

The 10th term of that particular A.P

SOLUTION :

Given that the zeroes of the polynomial $x^3+15x^2+66x+80$ are in A.P

First find the zeroes of the polynomial by using Synthetic Division

$x^3+15x^2+66x+80=0$

-5_|  1     15      66      80

        0    -5      -50     -80

      _________________

        1      10      16        0

∴ x+5 is a factor of the given polynomial

x+5=0

∴ x=-5 is a zero

Now $x^2+10x+16=0$

$(x+2)(x+8)=0$

x+2=0 or x+8=0

∴ x=-2 , x=-8

∴ x=-2,-5 and -8 are the zeroes

Since the zeroes are in A.P then we have

Let $a_1=-2$ , $a_2=-5$ and $a_3=-8$

Now the common difference $d=a_2-a_1$

Substitute the values we get

$d=-5-(-2)$

=-5+2

=-3

∴ d=-3

Now the common difference $d=a_3-a_2$

Substitute the values we get

$d=-8-(-5)$

=-8+5

=-3

∴ d=-3

Now find the 10th term

The general formula for A.P is

$a_n=a_1+(n-1)d$

Put n=10 , $a_1=-2$ and d=-3 in the formula we get

$a_{10}=-2+(10-1)(-3)$

$=-2+9(-3)$

$=-2-27$

∴ $a_{10}=-29$

The 10th term of that particular A.P is -29