Question

If ${{x}^{2}}+x+1$ is a factor of the polynomial $3{{x}^{3}}+8{{x}^{2}}+8x+3+5k$, then the value of k is
[a] 0
[b] $\dfrac{2}{5}$
[c] $\dfrac{5}{2}$
[d] -1

Answer 1

Hint: Divide $p\left( x \right)=3{{x}^{3}}+8{{x}^{2}}+8x+3+5k$ by ${{x}^{2}}+x+1$ and find the remainder obtained on division. Equate the remainder to 0 and hence form an equation in k. Solve for k. Hence find the value of k such that ${{x}^{2}}+x+1$ is a factor of p(x).

Complete step-by-step answer:
Alternatively, use the fact that ${{x}^{2}}+x+1=\left( x-\omega \right)\left( x-{{\omega }^{2}} \right)$, where $\omega $ is a complex cube root of unity and hence $x-\omega $ is a factor of p(x). Hence applying factor theorem conclude that $p\left( \omega \right)=0$ and hence find the value of k.

We have $p\left( x \right)=3{{x}^{3}}+8{{x}^{2}}+8x+3+5k=3{{x}^{3}}+3{{x}^{2}}+3x+5{{x}^{2}}+5x+3+5k$
Hence, we have
$p\left( x \right)=3x\left( {{x}^{2}}+x+1 \right)+5{{x}^{2}}+5x+3+5k$
Again, we have
$p\left( x \right)=3x\left( {{x}^{2}}+x+1 \right)+5{{x}^{2}}+5x+5+5k-2$
Hence, we have
$p\left( x \right)=3x\left( {{x}^{2}}+x+1 \right)+5\left( {{x}^{2}}+x+1 \right)+5k-2=\left( {{x}^{2}}+x+1 \right)\left( 3x+5 \right)+5k-2$
Hence remainder = 5k-2
But since ${{x}^{2}}+x+1$ is a factor of p(x), we have
5k-2 = 0
Adding 2 on both sides, we get
5k = 2
Dividing by 5 on both sides, we get
$k=\dfrac{2}{5}$
Hence option [b] is correct.

Note: Alternative solution:
We have ${{x}^{2}}+x+1=\left( x-\omega \right)\left( x-{{\omega }^{2}} \right)$, where $\omega $ is a complex cube root of unity.
Hence $x-\omega $ is a factor of p(x).
Hence, by factor theorem, we have
$p\left( \omega \right)=0$
Now, we have $p\left( x \right)=3{{x}^{3}}+8{{x}^{2}}+8x+3+5k$
Substituting $x=\omega $, we get
$p\left( \omega \right)=3{{\omega }^{3}}+8{{\omega }^{2}}+8\omega +3+5k$
We know that ${{\omega }^{3}}=1$
Hence, we have
$p\left( \omega \right)=3+8{{\omega }^{2}}+8\omega +3+5k=8\left( \omega +{{\omega }^{2}} \right)+6+5k$
We know that $1+\omega +{{\omega }^{2}}=0\Rightarrow \omega +{{\omega }^{2}}=-1$
Hence, we have
$p\left( \omega \right)=8\left( -1 \right)+6+5k=5k-2$
But $p\left( \omega \right)=0$
Hence, we have
$5k-2=0$
Hence, we have
$k=\dfrac{2}{5}$
Hence option [b] is correct.