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

If $x - \sqrt{5}$ is a factor of the cubic polynomial $x^{3} - 3 \sqrt{5} x^{2} + 13 x - 3 \sqrt{5}$ , then find all the zeroes of the polynomial.

Accepted Answer

Let $f (x) = x^{3} - 3 \sqrt{5} x^{2} + 13 x - 3 \sqrt{5}$ and given that $(x - \sqrt{5})$ is a one of the factor of $f (x)$ .

Now, using division algorithm,

$∴ x^{3} - 3 \sqrt{5} x^{2} + 13 x - 3 \sqrt{5} = (x^{2} - 2 \sqrt{5} x + 3) \times (x - \sqrt{5})$

[ $∵$ dividend $=$ divisor $\times$ quotient $+$ remainder]

$= (x - \sqrt{5}) [x^{2} - {(\sqrt{5} + \sqrt{2}) + (\sqrt{5} - \sqrt{2})} x + 3]$ [by splitting the middle term]

$= (x - \sqrt{5}) [x^{2} - (\sqrt{5} + \sqrt{2}) x - (\sqrt{5} - \sqrt{2}) x + (\sqrt{5} + \sqrt{2}) (\sqrt{5} - \sqrt{2})]$

$[∵ 3 = (\sqrt{5} + \sqrt{2}) (\sqrt{5} - \sqrt{2})]$

$= (x - \sqrt{5}) [x {x - (\sqrt{5} + \sqrt{2})} - (\sqrt{5} - \sqrt{2}) {x - (\sqrt{5} + \sqrt{2})}]$

$= (x - \sqrt{5}) {x - (\sqrt{5} + \sqrt{2})} {x - (\sqrt{5} - \sqrt{2})}$

Hence, all the zeroes of polynomial are $\sqrt{5}, (\sqrt{5} + \sqrt{2})$ and $(\sqrt{5} - \sqrt{2}) .$

If $x - \sqrt{5}$ is a factor of the cubic polynomial $x^{3} - 3 \sqrt{5} x^{2} + 13 x - 3 \sqrt{5}$ , then find all the zeroes of the polynomial.

Important Questions on Polynomials

Learn and Prepare for any exam you want !

If x-5 is a factor of the cubic polynomial x3−35x2+13x−35, then find all the zeroes of the polynomial.

Important Questions on Polynomials

Learn and Prepare for any exam you want !

If $x - \sqrt{5}$ is a factor of the cubic polynomial $x^{3} - 3 \sqrt{5} x^{2} + 13 x - 3 \sqrt{5}$ , then find all the zeroes of the polynomial.