Question

If $x^4-5x^3+9x^2-7x+2=0$ has a multiple root of order $3$ then the roots are:

• A. $1,1,1,2$
• B. $1,2,2,2$
• C. $-1,-1,-1,2$
• D. $1,-2,-2,-2$

Answer 1

The correct option is A $1,1,1,2$

Let the root will multiplicity $3=a$ & other root $=b$

So, we can write the polynomial as product of ${(x-a)}^3(x-b)$

$\therefore$ $x^4-5x^3+9x^2-7x+2=(x^3-3x^{2}a+3x{a}^2-a^3)(x-b)$

$=x^4-{(b+3a)x}^3+(3ab+3a^2)x^2-(3a^{2}b+a^3)x+a^{3}b$

Comparing coefficients on both sides.

$\Rightarrow b+3a=5⟶\left(1\right)$

$\Rightarrow a^{3}b=2⟶\left(2\right)$

Solving two equations,

$b=\frac{2}{a^3}\Rightarrow \frac{2}{a^3}+3a-5\Rightarrow 2+3a^6=5a^3$

$\Rightarrow 3a^6-5a^3=-2$

$\Rightarrow a^3(3a^3-5)=-2\Rightarrow 1^3({3a}^3-5)=(1\times 2)$

$\Rightarrow a=1$

$\therefore b=\frac{2}{1^3}=\frac{2}{1}=2$

$\therefore$ $a=1$, $b=2$

So, the option, $(1,1,1,2)$ is correct.