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Dividing a Polynomial by a Polynomial

One of the ze...

Question

One of the zeroes of cubic polynomial $(3 x^{3} - 5 x^{2} - 11 x - 3)$ is $\frac{- 1}{3}$ . Find the other zeroes.

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Solution

Since it is given $(x - \frac{- 1}{3})$ is one factor of the given polynomial, by division algorithm we can get the quotient as $(3 x^{2} - 6 x - 9) .$

To find the zeroes of $(3 x^{2} - 6 x - 9)$ , we can factorize it.

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

So, the other zeroes are 3 and -1.

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Similar questions

Q. lf one zero of the cubic polynomial

P (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

\frac{- 1}{3}

, then other two zeroes are ________.

- \frac{1}{3}

is the zero of the cubic polynomial

f (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

the other zeros are

3, - 1, \frac{- 1}{3}

are the zeros of the cubic polynomial

3 x^{3} - 5 x^{2} - 11 x - 3

a, - 1, - \frac{1}{a}

are the zeroes of the cubic polynomial

p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

3, - 1, - \frac{1}{3}

are the zeroes of the cubic polynomial

p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

and then verify the relationship between the zeroes and the coefficients

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