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Question
If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then the product of the other two zeroes is ______.
Options
b – a + 1
b – a – 1
a – b + 1
a – b –1
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Solution
If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then the product of the other two zeroes is b – a + 1.
Explanation:
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,
Product of all zeroes = `(-1)^3`
`"Constant term"/("Coefficient of" x^3) = - c/1`
αβγ = – c
⇒ (–1)βγ = −c .......[∴ α = –1]
⇒ βγ = c
⇒ βγ = 1 – a + b ......[From equation (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,
`x^2 + (a - 1)x + (b - a + 1)`
`x + 1")"overline(x^3 + ax^2 + bx + c)`
x3 + x2
(a – 1)x2 + bx
(a – 1)x2 + (a – 1)x
(b – a + 1)x + c
(b – a + 1)x (b – a + 1)
(c – b + a – 1)
⇒ 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
Product of all zeroes = `(-1)alpha*beta`
= `(-"Constant term")/("Coefficient of" x^3)`
⇒ `- alpha*beta = (-(b - a + 1))/1`
⇒ `alpha beta` = – a + b + 1
Hence the required product of other two roots is (–a + b + 1).
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