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For what value of $k$, is the polynomial $f(x)\ =\ 3x^4\ -\ 9x^3\ +\ x^2\ +\ 15x\ +\ k$ completely divisible by $3x^2\ -\ 5$?
Given:
Given polynomial is $f(x)\ =\ 3x^4\ -\ 9x^3\ +\ x^2\ +\ 15x\ +\ k$.
The divisor is $3x^2\ -\ 5$.
To do:
We have to find the value of $k$.
Solution:
If $f(x)\ =\ 3x^4\ -\ 9x^3\ +\ x^2\ +\ 15x\ +\ k$ is completely divisible by $3x^2\ -\ 5$ then it is a factor of $f(x)$.
To find the zeroes, put $3x^2\ -\ 5 = 0$
$3x^2-5=0$
$3x^2=5$
$x^2=\frac{5}{3}$
$x=\sqrt{\frac{5}{3}}$ or $x=-\sqrt{\frac{5}{3}}$
$x=\sqrt{\frac{5}{3}}$ is a root of $f(x)$.
Therefore,
$f(\sqrt{\frac{5}{3}}) = 3(\sqrt{\frac{5}{3}})^4-9(\sqrt{\frac{5}{3}})^3+(\sqrt{\frac{5}{3}})^2+15(\sqrt{\frac{5}{3}})+k=0$
$ \begin{array}{l}
3\left(\sqrt{\frac{5}{3}}\right)^{4} -9\left(\sqrt{\frac{5}{3}}\right)^{3} +\left(\sqrt{\frac{5}{3}}\right)^{2} +15\left(\sqrt{\frac{5}{3}}\right) +k=0\\
\\
3\left(\frac{5}{3}\right)^{2} -9\left(\frac{5}{3}\right)\left(\sqrt{\frac{5}{3}}\right) +\frac{5}{3} +15\left(\sqrt{\frac{5}{3}}\right) +k=0\\
\\
\frac{25}{3} -15\left(\sqrt{\frac{5}{3}}\right) +\frac{5}{3} +15\left(\sqrt{\frac{5}{3}}\right) +k=0\\
\\
\frac{30}{3} +k=0\\
\\
10+k=0\\
\\
k=-10
\end{array}$
The value of $k$ is $-10$.
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