Question

2 is a zero of the polynomial function \[ f(x)=x^4+kx^3+22x^2-6x-20. \] If \(-2,\alpha,\beta\) are the roots of the equation \[ x^3+3x^2+2kx-40=0 \] and \(\alpha<\beta\), then \(2\alpha+3\beta=\)

• A. \(1\)
• B. \(2\)
• C. \(5\)
• D. \(8\)

Hint:

Whenever a root of a polynomial is given, immediately substitute it into the polynomial to determine unknown parameters. After finding the parameter, factor the equation completely before applying Vieta's formulas.

Answer 1

The Correct Option is B

Concept: This problem combines two important algebraic ideas:

• If a number is a zero of a polynomial, then substitution makes the polynomial equal to zero.

• For a cubic equation, Vieta's relations connect the roots with the coefficients.
We shall first determine the value of \(k\) using the given polynomial and then use Vieta's formulas to determine \(\alpha\) and \(\beta\).

Step 1: Use the fact that \(2\) is a zero of \(f(x)\) Since \(2\) is a root, \[ f(2)=0. \] Substituting \(x=2\), \[ 2^4+k(2^3)+22(2^2)-6(2)-20=0. \] \[ 16+8k+88-12-20=0. \] \[ 72+8k=0. \] \[ 8k=-72. \] \[ k=-9. \] Thus, \[ \boxed{k=-9}. \]

Step 2: Substitute \(k=-9\) into the cubic equation The cubic becomes \[ x^3+3x^2-18x-40=0. \] We are told that \(-2\) is one of its roots. Hence, \[ (x+2) \] must be a factor.

Step 3: Divide the cubic by \((x+2)\) Using synthetic division: \[ \begin{array}{r|rrrr} -2 & 1 & 3 & -18 & -40\\ & & -2 & -2 & 40\\ \hline & 1 & 1 & -20 & 0 \end{array} \] Therefore, \[ x^3+3x^2-18x-40 = (x+2)(x^2+x-20). \] Now factor the quadratic: \[ x^2+x-20 = (x+5)(x-4). \] Hence the roots are \[ -2,\,-5,\;4. \]

Step 4: Identify \(\alpha\) and \(\beta\) Given \[ \alpha<\beta, \] the remaining roots are \[ \alpha=-5, \qquad \beta=4. \]

Step 5: Compute \(2\alpha+3\beta\) Substituting the values, \[ 2\alpha+3\beta = 2(-5)+3(4). \] \[ =-10+12. \] \[ =2. \] Therefore, \[ \boxed{2\alpha+3\beta=2}. \] Hence the correct answer is \[ \boxed{(B)\;2}. \]