9. If one of the zeroes of cubic
polynomial is x^3+ax^2+bx+c is
--1, then product of other two
zeroes is:
Answers
Answer:-
(a) Let p(x) = x3 + ax2 + bx + c
(a) Let p(x) = x3 + ax2 + bx + cLet a, p and y be the zeroes of the given cubic polynomial p(x).
(a) Let p(x) = x3 + ax2 + bx + cLet a, p and y be the zeroes of the given cubic polynomial p(x).∴ α = -1 [given]
(a) Let p(x) = x3 + ax2 + bx + cLet a, p and y be the zeroes of the given cubic polynomial p(x).∴ α = -1 [given]and p(−1) = 0
(a) Let p(x) = x3 + ax2 + bx + cLet 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
(a) Let p(x) = x3 + ax2 + bx + cLet 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
(a) Let p(x) = x3 + ax2 + bx + cLet 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,
We know that,αβγ = -c
We know that,αβγ = -c⇒ (-1)βγ = −c [∴α = -1]
We know that,αβγ = -c⇒ (-1)βγ = −c [∴α = -1]⇒ βγ = c
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.
product of the other two roots is 1 -a + b.Alternate Method
product of the other two roots is 1 -a + b.Alternate MethodSince, -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).
product of the other two roots is 1 -a + b.Alternate MethodSince, -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,
product of the other two roots is 1 -a + b.Alternate MethodSince, -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)
product of the other two roots is 1 -a + b.Alternate MethodSince, -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)}
product of the other two roots is 1 -a + b.Alternate MethodSince, -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
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