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

Answer 1

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).