If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then the product of the other two zeroes is ______.

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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). $\therefore$ $\alpha$ = –1 ......[Given] And p(−1) = 0 $\Rightarrow$ (–1)3 + a(–1)2 + b(–1) + c = 0 $\Rightarrow$ –1 + a – b + c = 0 $\Rightarrow$ c = 1 – a + b ......(i) We know that, Product of all zeroes = (-1)^3 `"Constant term"/("Coefficient of" x^3) = - c/1` $\alpha$$\beta$γ = – c $\Rightarrow$ (–1)$\beta$γ = −c .......[$\therefore$ $\alpha$ = –1] $\Rightarrow$ $\beta$γ = c $\Rightarrow$ $\beta$γ = 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) $\Rightarrow$ x3 + ax2 + bx + c = (x + 1)x {x2 + (a – 1)x + (b – a + 1) > + (c – b + a – 1) $\Rightarrow$ 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) $\Rightarrow$ - alpha*beta = (-(b - a + 1))/1 $\Rightarrow$ alpha beta` = – a + b + 1 Hence the required product of other two roots is (–a + b + 1).

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