Question

If \(x^2 + px + 1\) is a factor of \(ax^3 + bx + c\), then:

• A. \(a^2 + c^2 = ab + 3\)
• B. \(a^2 - c^2 = ab\)
• C. \(a^2 - c^2 = -ab\)
• D. \(a^2 + c^2 = ab\)

Hint:

When a quadratic is a factor of a cubic, assume the quotient is linear, expand, and equate coefficients to find relations among coefficients.

Answer 1

The Correct Option is B

Step 1: Match the leading coefficient directly.
Since \(x^2+px+1\) divides \(ax^3+bx+c\) (which has no \(x^2\) term), write \(ax^3+bx+c=(x^2+px+1)(ax+q)\); the quotient must start with \(ax\) so that the \(x^3\) coefficient works out to \(a\).

Step 2: Expand and match coefficients.
\((x^2+px+1)(ax+q)=ax^3+(q+ap)x^2+(pq+a)x+q\)
Matching with \(ax^3+0\cdot x^2+bx+c\): \(q+ap=0\), \(pq+a=b\), \(q=c\).

Step 3: Eliminate q and p.
From \(q=c\) and \(q+ap=0\), we get \(p=-\frac{c}{a}\).
Substituting into \(b=pq+a\): \(b=\left(-\frac{c}{a}\right)c+a=a-\frac{c^2}{a}=\frac{a^2-c^2}{a}\)

Step 4: Conclusion.
Multiplying through by \(a\) gives \(ab=a^2-c^2\).
\[ \boxed{a^2-c^2=ab} \]