If one of the zeroes of the cubic polynomial x3+ ax2+ bx + c is –
1, then the
product of the other two zeroes is
Answers
Answer:
Step-by-step explanation:
(a) Let p(x) = x3 + ax2 + bx + c
Let a, p and y 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 …(i)
We know that,
αβγ = -c
⇒ (-1)βγ = −c [∴α = -1]
⇒ βγ = c
⇒ βγ = 1 -a + b [from Eq. (i)]
Hence, product of the other two roots is 1 -a + b.
Alternate Method
Since, -1 is one of the zeroes of the cubic polynomial f(x) = x2 + ax2 + bx + c i.e., (x + 1) is a factor of f{x).
Now, using division algorithm,
⇒x3 + ax2 + bx +c = (x + 1) x {x2 + (a – 1)x + (b – a + 1)> + (c – b + a -1)
⇒x3 + ax2 + bx + (b – a + 1) = (x + 1) {x2 + (a – 1)x + (b -a+ 1)}
Let a and p be the other two zeroes of the given polynomial, then