Question

The order of multiple roots of 2 of the equation ${{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}+4x-8=0$ is
[a] 1
[b] 2
[c] 3
[d] 4

Answer 1

Hint: Use the fact that if the multiplicity of a root x= a is r in the polynomial p(x), then there exists a polynomial g(x) such that $p\left( x \right)={{\left( x-a \right)}^{r}}g\left( x \right)$ and $g\left( a \right)\ne 0$. Use factor theorem to check that if x-2 is a factor of $p\left( x \right)={{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}+4x-8$. If yes, then write $p\left( x \right)=\left( x-2 \right)g\left( x \right)$ using the long-division method. Apply the same process on g(x). Continue till g(x) is no more divisible by x-2. Hence find the order of the root 2 in the equation ${{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}+4x-8=0$

Complete step-by-step answer:
Let $p\left( x \right)={{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}+4x-8$
Now, we have
$p\left( 2 \right)={{2}^{4}}-5\times {{2}^{3}}+6\times {{2}^{2}}+4\times 2-8=16-40+24+8-8=0$
Hence by factor theorem, x-2 is a factor of p(x).
Dividing p(x) by x-2 using long-division method shown as follows:

Hence, we have
$p\left( x \right)=\left( x-2 \right)\left( {{x}^{3}}-3{{x}^{2}}+4 \right)$
Now, we have $g\left( x \right)={{x}^{3}}-3{{x}^{2}}+4$
Now, we have
$g\left( 2 \right)={{2}^{3}}-3\times {{2}^{2}}+4=8-12+4=0$
Hence by factor theorem, x-2 is a factor of g(x).
Hence, we have

Hence, we have
$\begin{align}
  & g\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}-x-2 \right) \\
 & \Rightarrow p\left( x \right)={{\left( x-2 \right)}^{2}}\left( {{x}^{2}}-x-2 \right) \\
\end{align}$
Let $h\left( x \right)={{x}^{2}}-x-2$
Hence, we have
$h\left( 2 \right)={{2}^{2}}-2-2=0$
Hence by factor theorem, x-2 is a factor of h(x)
Hence, we have

Hence, we have
$\begin{align}
  & h\left( x \right)=\left( x-2 \right)\left( x+1 \right) \\
 & \Rightarrow p\left( x \right)={{\left( x-2 \right)}^{3}}\left( x+1 \right) \\
\end{align}$
Let f(x) = x+1
Since $f\left( 2 \right)=1+2\ne 0$, we have
3 is the order of the root 2 of the equation ${{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}+4x-8=0$
Hence option [c] is correct.

Note: Alternative solution: Best method:
We know that if r is the multiplicity of root x=a in the polynomial p(x), then $p\left( a \right),{{\left. \dfrac{d}{dx}p\left( x \right) \right|}_{x=a}},{{\left. \dfrac{{{d}^{2}}}{d{{x}^{2}}}p\left( x \right) \right|}_{x=a}},\cdots ,{{\left. \dfrac{{{d}^{r-1}}}{d{{x}^{r-1}}}p\left( x \right) \right|}_{x=r}}$ are all 0.
Now, we have
$p\left( x \right)={{x}^{4}}-5{{x}^{3}}+6{{x}^{2}}+4x-8$
Hence, we have
$p\left( 2 \right)={{2}^{4}}-5\times {{2}^{3}}+6\times {{2}^{2}}+4\,\times 2-8=16-40+24+8-8=0$
Now, we have
$\dfrac{d}{dx}p\left( x \right)=4{{x}^{3}}-15{{x}^{2}}+12x+4$
Hence, we have
${{\left. \dfrac{d}{dx}p\left( x \right) \right|}_{x=2}}=4\times {{2}^{3}}-15\times {{2}^{2}}+12\times 2+4=32-60+24+4=0$
Now, we have
$\dfrac{{{d}^{2}}}{d{{x}^{2}}}p\left( x \right)=12{{x}^{2}}-30x+12$
Hence, we have
${{\left. \dfrac{{{d}^{2}}}{d{{x}^{2}}}p\left( x \right) \right|}_{x=2}}=12\times {{2}^{2}}-30\times 2+12=48-60+12=0$
Now, we have
$\dfrac{{{d}^{3}}}{d{{x}^{3}}}p\left( x \right)=24x-30$
Hence, we have
${{\left. \dfrac{{{d}^{3}}}{d{{x}^{3}}}p\left( x \right) \right|}_{x=2}}=24\times 2-30=48-30=18\ne 0$
Hence, the multiplicity of the root x= 2 in p(x) is 2+1 =3.
Hence option [c] is correct.