Name: Division of a Trinomial by a Trinomial
Uploaded: 2019-07-19
Duration: 3 min 36 s
Description: Can we divide a polynomial by another polynomial to get the quotient polynomial and the remainder polynomial? Let's explore this concept in this video.

Question

Find all the zeroes of

2 x^{4} - 3 x^{3} - 3 x^{2} - 6 x - 2

, if we have two of its zeroes as

\sqrt{2}

and

- \sqrt{2}

.

Accepted Answer

Given:

Polynomial $2 x^{4} - 3 x^{3} - 3 x^{2} - 6 x - 2$ has two of its zeroes as $\sqrt{2}$ and $- \sqrt{2}$ .

We can say that $(x - \sqrt{2})$ and $(x + \sqrt{2})$ are factors of the polynomial.

So, $(x - \sqrt{2}) (x + \sqrt{2}) i . e (x^{2} - 2)$ is also a factor of the polynomial.

Let us perform the division as shown below:

$x^{2} - 2) 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 (2 x^{2} - 3 x + 1 2 x^{4} - 4 x^{2} - +________________0 - 3 x^{3} + x^{2} + 6 x - 2 - 3 x^{3} + 6 x + -________________0 + x^{2} + 0 - 2 + x^{2} - 2 - +________________0$

So, other factor of the polynomial is $2 x^{2} - 3 x + 1$ .

From middle term factorization, we have $2 x^{2} - 3 x + 1 = 2 x^{2} - 2 x - x + 1$

$= (2 x - 1) (x - 1)$

$\Rightarrow x = \frac{1}{2}, x = 1$

Hence, all zeroes are $\sqrt{2}, - \sqrt{2}, \frac{1}{2}, 1$ .

Find all the zeroes of $2 x^{4} - 3 x^{3} - 3 x^{2} - 6 x - 2$ , if we have two of its zeroes as $\sqrt{2}$ and $- \sqrt{2}$ .

Important Questions on Polynomials

Learn and Prepare for any exam you want !

Find all the zeroes of 2x4-3x3-3x2-6x-2, if we have two of its zeroes as 2 and -2.

Important Questions on Polynomials

Learn and Prepare for any exam you want !

Find all the zeroes of $2 x^{4} - 3 x^{3} - 3 x^{2} - 6 x - 2$ , if we have two of its zeroes as $\sqrt{2}$ and $- \sqrt{2}$ .