Question

If x – 2 is a factor of the polynomial \(x^3-6x^2+ax-8,\) , then the value of a is

• A. 10
• B. 12
• C. 14
• D. 18

Answer 1

The Correct Option is B

We are given that \( x - 2 \) is a factor of the polynomial:

\[ P(x) = x^3 - 6x^2 + ax - 8 \] 

Step 1: Factor Theorem

Since \( x - 2 \) is a factor, substituting \( x = 2 \) into \( P(x) \) must yield 0:

\[ P(2) = (2)^3 - 6(2)^2 + a(2) - 8 = 0 \]

Step 2: Solving for \( a \)

\[ 8 - 24 + 2a - 8 = 0 \]

\[ 2a - 24 + 8 - 8 = 0 \]

\[ 2a - 24 = 0 \]

\[ 2a = 24 \]

\[ a = 12 \]

Final Answer: \( a = 12 \).

Answer 2

If \(x – 2\) is a factor of the polynomial \(f(x)=x^3-6x^2+ax-8\), then by the Factor Theorem, \(f(2) = 0\). 
Let's calculate \(f(2)\):
Substitute \(x = 2\) into the polynomial:

\(f(2)=2^3-6(2)^2+a(2)-8\)

This simplifies to:

\(= 8 - 6(4) + 2a - 8\)

Evaluate each term:

\(= 8 - 24 + 2a - 8\)

\(= -24 + 2a\)

Setting \(f(2) = 0\) gives:

\(-24 + 2a = 0\)

Solve for \(a\)

\(2a = 24\)

Divide both sides by 2:

\(a = 12\)

The correct value of \(a\) is 12.