Obtain all other zeroes of the polynomial $9 x^{4} - 6 x^{3} - 35 x^{2} + 24 x - 4$ , if two of its zeroes are $2$ and $- 2$ .

Open in App

Solution

It is given that $2$ and $- 2$ are zeroes of the polynomial $9 x^{4} - 6 x^{3} - 35 x^{2} + 24 x - 4$ that is:

$(x + 2) (x - 2) = x^{2} - 4$

Therefore, we divide the polynomial $9 x^{4} - 6 x^{3} - 35 x^{2} + 24 x - 4$ by $x^{2} - 4$ as shown in the above image:

From the above division, we conclude that the quotient is $9 x^{2} - 6 x + 1$ and the remainder is $0$ .

Now, let us factorize the quotient $9 x^{2} - 6 x + 1$ to find the other zereos of the given polynomial:

$9 x^{2} - 6 x + 1 = 0 \Rightarrow 9 x^{2} - 3 x - 3 x + 1 = 0 \Rightarrow 3 x (3 x - 1) - 1 (3 x - 1) = 0 \Rightarrow (3 x - 1) (3 x - 1) = 0 \Rightarrow (3 x - 1) = 0, (3 x - 1) = 0 \Rightarrow x = \frac{1}{3}, x = \frac{1}{3}$

Hence, the other zeroes of the polynomial $9 x^{4} - 6 x^{3} - 35 x^{2} + 24 x - 4$ are $\frac{1}{3}, \frac{1}{3}$ .

Suggest Corrections

Similar questions

2 x^{4} + 3 x^{3} - 15 x^{2} - 24 x - 8

2 \sqrt{2}

- 2 \sqrt{2}

\sqrt{2}

6 x^{3} + \sqrt{2} x^{2} - 10 x - 4 \sqrt{2}

x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35

2 \pm \sqrt{3}

3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5

\sqrt{\frac{5}{3}}

- \sqrt{\frac{5}{3}}

3 x^{4} + 6 x^{3} - 2 x^{2} - 10 x - 5

\sqrt{\frac{5}{3}} a n d - \sqrt{\frac{5}{3}}

Join BYJU'S Learning Program