Question

If one of the zeroes of the cubic polynomial + ax^3 + bx^2 + cx + d is -1, then the
product of the other two zeroes is
(A) b-a +1 (B) b-a-1 (C) a-b+1 (D)a-b-1​

Answer 1

A. Using Factor Theorem

B. Finding two Factors

C. Relation between Roots and Coefficients

A

One of the zero is -1 so one factor will be x+1.

After substituting x=-1

$\sf{a-b+c-d=0}$

$\sf{\therefore{d=a-b+c}}$ ...(1)

B

To find another factor, we divide by x+1.

$\sf{ax^3+bx^2+cx+d=(x+1)\{ax^2-(a-b)x+a-b+c\}}$

Now all zeros will come from:

• $\sf{(x+1)=0}$
• $\sf{\{ax^2-(a-b)x+a-b+c\}=0}$ ...(2)

C

According to the relation between roots and coefficients,

the product of the other two zeros is $\sf{\alpha \beta =\dfrac{c}{a} }$.

$\sf{\therefore {\dfrac{a-b+c}{a} =1-\dfrac{b-c}{a} }}$

or

$\sf{\therefore{\dfrac{a-b+c}{a} =\dfrac{d}{a} }}$