Question

If the zeros of the polynomial f(x) = 2x³ − 15x² + 37x − 30 are in A.P., find them.

Answer 1

Let α = a - d, β = a and γ = a + d be the zeros of the polynomial

f(x) = 2x³ − 15x² + 37x − 30

Therefore

`alpha+beta+gamma=("coefficient of "x^2)/("coefficient of "x^3)`

`=-((-15)/2)`

`=15/2`

 

`alphabetagamma="-constant term"/("coefficient of "x^2)`

`=-((-30)/2)`

= 15

Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`

`(a-d)+a+(a+d)=15/2`

`a+a+a-d+d=15/2`

`3a=15/2`

`a=15/2xx1/3`

`a=5/2`

Product of the zeroes `="-constant term"/("coefficient of "x^2)`

`alphabetagamma=15`

`(a-d)+a+(a+d)=15`

`a(a^2-d^2)=15`

Substituting a = 5/2 we get

`5/2((5/2)^2-d^2)=15`

`5/2(25/4-d^2)=15`

`25/4-d^2=15xx2/5`

`25/4-d^2=3xx2`

`25/4-d^2=6`

`-d^2=6-25/4`

`-d^2=(24-25)/4`

`-d^2=(-1)/4`

`d^2=1/4`

`d xx d=1/2xx1/2`

`d=1/2`

Therefore, substituting a=5/2 and d=1/2 in α = a - d, β = a and γ = a + d

α = a - d

`alpha=5/2-1/2`

`alpha=(5-1)/2`

`alpha=4/2`

`alpha=2`

 

β = a

`beta=5/2`

 

γ = a + d

`gamma=5/2+1/2`

`gamma=(5+1)/2`

`gamma=6/2`

`gamma=3`

Hence, the zeros of the polynomial are 2, `5/2` , 3.