Let p(x)=x3+ax2+bx+c
Let α β and γ be the zeroes of the given cubic polynomial p (x)
∴ α=−1 [given]
and p( - 1) = 0
⇒ (−1)3+a(−1)2+b(−1)+c=0
⇒ −1+a−b+c=0
⇒ c=1–a+b
We know that,
Product of the zeroes = (−1)3constant termcoefficient of x3=−c1
αβγ=−c
⇒ (−1)βγ=−c [∵α=−1]
⇒βγ=c
⇒ βγ=1−a+b [fromEq.(i)]
Hence, product of the other two roots is 1 – a + b