Question:
If \(\alpha\) is a root of the equation
\[
x^4-4x^3+16x-16=0
\]
of multiplicity \(m\), then
\[
m^2-4m+2=
\]
If \(\alpha\) is a root of the equation
\[
x^4-4x^3+16x-16=0
\]
of multiplicity \(m\), then
\[
m^2-4m+2=
\]
Show Hint
To determine the multiplicity of a root, completely factorize the polynomial. The exponent of the factor \((x-\alpha)\) gives the multiplicity of the root \(\alpha\).
Updated On: Jul 9, 2026
- \(-1\)
- \(1\)
- \(0\)
- \(2\) \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
A root \(\alpha\) of multiplicity \(m\) satisfies
\[
(x-\alpha)^m
\]
as a factor of the polynomial.
To find the multiplicity, factorize the polynomial completely.
Step 1: Factorize the given polynomial. Given \[ P(x)=x^4-4x^3+16x-16. \] Grouping terms, \[ P(x) = x^3(x-4)+4(x-4). \] Taking the common factor \((x-4)\), \[ P(x) = (x-4)(x^3+4). \] Using \[ a^3+b^3=(a+b)(a^2-ab+b^2), \] we get \[ x^3+4 = x^3+2^3 = (x+2)(x^2-2x+4). \] Hence, \[ P(x) = (x-4)(x+2)(x^2-2x+4). \]
Step 2: Check the multiplicity of the roots. The factors \[ (x-4),\quad (x+2),\quad (x^2-2x+4) \] all occur only once. Therefore, every root of the equation has multiplicity \[ m=1. \]
Step 3: Evaluate \(m^2-4m+2\). Substituting \(m=1\), \[ m^2-4m+2 = 1^2-4(1)+2. \] \[ =1-4+2. \] \[ =-1. \]
Step 4: Write the final answer. \[ \boxed{-1} \]
Step 1: Factorize the given polynomial. Given \[ P(x)=x^4-4x^3+16x-16. \] Grouping terms, \[ P(x) = x^3(x-4)+4(x-4). \] Taking the common factor \((x-4)\), \[ P(x) = (x-4)(x^3+4). \] Using \[ a^3+b^3=(a+b)(a^2-ab+b^2), \] we get \[ x^3+4 = x^3+2^3 = (x+2)(x^2-2x+4). \] Hence, \[ P(x) = (x-4)(x+2)(x^2-2x+4). \]
Step 2: Check the multiplicity of the roots. The factors \[ (x-4),\quad (x+2),\quad (x^2-2x+4) \] all occur only once. Therefore, every root of the equation has multiplicity \[ m=1. \]
Step 3: Evaluate \(m^2-4m+2\). Substituting \(m=1\), \[ m^2-4m+2 = 1^2-4(1)+2. \] \[ =1-4+2. \] \[ =-1. \]
Step 4: Write the final answer. \[ \boxed{-1} \]
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