Question

If the polynomial f(x) =3x4-9x3+x^2+15x+k is completely divisible by 3x2-5 find the value of k and hence the other two zero of the polynomial

Answer 1

value of k is -10
zeroes are 1, 2, (-√5/√3), (√5/√3)

Answer 2

k= -10 is the value of k and the other two zero of the polynomial are 1 and 2.

Given:

$f(x)=3 x^{4}-9 x^{3}+x^{2}+15 x+k$

To find:

Value of k and other two zero of the polynomial = ?

Solution:  

The given polynomial is  $f(x)=3 x^{4}-9 x^{3}+x^{2}+15 x+k$

We need to find the value of k and the other roots ( zeroes) of the polynomial.

Also given that f(x)is divisible by $3 x^{2}-5$

when the polynomial is divided by $3 x^{2}-5$ the remainder is 0.  

Let us divide f(x)with $3 x^{2}-5$ is attached below

$3 x^{2}-5$  and $x^{2}-3 x+2$ are the roots of the given polynomial.  

That is ,  $3 x^{2}-5=0$ $\Rightarrow(\sqrt{3 x}-\sqrt{5})(\sqrt{3 x}+\sqrt{5})=0$

$x=\frac{\sqrt{5}}{\sqrt{3}},-\frac{\sqrt{5}}{\sqrt{3}}$

also, $x^{2}-3 x+2=0$

$\begin{array}{l}{x^{2}-2 x-x+2=0} \\ {x(x-2)-1(x-2)=0} \\ {(x-1)(x-2)=0}\end{array}x=1 \text { or } x=2$

Therefore, $\bold{x=1,2, \frac{\sqrt{5}}{\sqrt{3}},-\frac{\sqrt{5}}{\sqrt{3}}}$ are the roots of the given polynomial.