Question

show that -1,1 and 3 are all zeros of the polynomial x^3-3x^2-x+3​

Answer 1

Step-by-step explanation:

For 1

st

AP

a=8,d=20

For 2

nd

AP

a

′

=−30,d

′

=8

According to the question, S

n

=S

2n

′

⇒

2

n

[2a+(n−1)d]=

2

2n

[2a

′

+(2n−1)d

′

]

⇒[2(8)+(n−1)20]=2[2(−30)+(2n−1)8]

⇒2×8+n×20−20=2[−60+16n−8]

⇒20n−4=−136+32n

⇒−32n+20n=−136+4

⇒n=11