Question

lf one zero of the cubic polynomial $P\left(x\right)=3x^3-5x^2-11x-3$ is $\frac{-1}{3}$, then other two zeroes are ________.

• A. $-3,1$
• B. $1,3$
• C. $-1,3$
• D. Both (A) and (C)

Answer 1

Answer: (C)

$-1,3$

$-1/3$ is a zero of the polynomial,

$\Rightarrow (x+\frac{1}{3})$ will be one of the factors of the polynomial

Using long division method (refer attache image) we know the quotient $=(3x^2-6x-9)$

Factorizing the quotient,

$(3x^2-6x-9)=(3x^2-9x+3x-9)=3x(x-3)+3(x-3)=3(x+1)(x-3)$

From the above equation we can say, the other 2 zeroes of the polynomial are $x=-1,3$