If (x² + px + 1) is a factor of (ax³ + bx + c), then (A) a² + c² = -ab (B) a² - c² = -ab (C) a² - c²

Q: Wait, why do we assume the quotient is $(ax + c)$? How do we know it's not just $(mx + n)$?

That is a great observation! We actually do start with $(mx + n)$. However, by comparing the leading term $x^2 \cdot mx = ax^3$, we immediately see $m = a$. Similarly, comparing the constant term $1 \cdot n = c$, we get $n = c$. So, $(ax + c)$ is not an assumption; it is the only possible linear factor that satisfies the identity.

Q: Why did you add $0x^2$ to the cubic polynomial? Is that allowed?

Absolutely! In polynomial identity problems, we are comparing two polynomials. For them to be identical, the coefficients of every power of $x$ must match. Since the original cubic $ax^3 + bx + c$ has no $x^2$ term, its coefficient is implicitly zero. Writing it as $ax^3 + 0x^2 + bx + c$ is a powerful tool to ensure we don't miss any equations during the comparison step.

Q: How do I know which coefficient to compare first? Does it matter?

It doesn't strictly matter, but it is strategic. Comparing the $x^2$ coefficient gave us $0 = c + ap$, which allowed us to isolate $p$ easily. If we had started with the $x$ coefficient, we would have had an equation with both $b$ and $p$, which is harder to solve. Always look for the 'simplest' equation first to unlock the variables!

Question

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

Options:

(A) $a^2 + c^2 = -ab$

(B) $a^2 - c^2 = -ab$

(C) $a^2 - c^2 = ab$

(D) none of these

Sigma Percentile · Accepted Answer

### The Symphony of Coefficients: Unlocking Polynomial Identities

Imagine you are standing before a grand, complex structure—a cubic polynomial, $f(x) = ax^3 + bx + c$. It seems solid, but the problem whispers a secret: it is built from smaller, simpler blocks.

Specifically, it contains a quadratic factor, $g(x) = x^2 + px + 1$. In the world of algebra, when we say one polynomial is a factor of another, we are essentially saying that the cubic is a product of the quadratic and something else. This is the fundamental key to our journey.

### Phase 1: The Detective's Intuition

Since our cubic polynomial has a degree of $3$ and our quadratic factor has a degree of $2$, the missing piece—the quotient—must be a linear polynomial. Let us call it $q(x) = mx + n$.

We can use the 'leading term' and 'constant term' logic to find these coefficients. Look at the highest power: $x^2$ multiplied by $mx$ must yield $ax^3$. This forces $m = a$.

Now look at the constant: $1$ multiplied by $n$ must yield $c$. Thus, $n = c$. Just like that, we have identified our linear factor: $q(x) = ax + c$.

### Phase 2: The Identity Principle

Now, we set up our identity: 
$$ax^3 + bx + c = (x^2 + px + 1)(ax + c)$$

To make our comparison foolproof, let us rewrite the left side to explicitly show the missing $x^2$ term: $ax^3 + 0x^2 + bx + c$. This $0x^2$ is our secret weapon. It acts as a placeholder, ensuring that when we expand the right side, we have a clear target for every power of $x$.

### Phase 3: The Algebraic Dance

Let us expand the right side: $(x^2 + px + 1)(ax + c)$. Multiplying this out, we get:
$$ax^3 + cx^2 + apx^2 + cpx + ax + c$$

Grouping the terms by their powers of $x$, we obtain:
$$ax^3 + (c + ap)x^2 + (cp + a)x + c$$

Because the two polynomials are identical, their coefficients must be equal. Comparing the $x^2$ coefficients, we get $0 = c + ap$, which leads us to:
$$p = -\frac{c}{a}$$

Next, we compare the $x$ coefficients: $b = cp + a$. Substituting our value of $p$ into this equation, we get:
$$b = c\left(-\frac{c}{a}\right) + a$$

Multiplying the entire equation by $a$ to clear the denominator, we arrive at $ab = -c^2 + a^2$. Rearranging this, we find the beautiful, final relation:
**$$a^2 - c^2 = ab$$**

### Conclusion

We started with a daunting cubic expression and, through the simple, rhythmic application of the identity principle, stripped away the complexity to reveal a clean, elegant relationship between the coefficients.

This is the heart of JEE mathematics—not just solving for $x$, but understanding the structural harmony of the equations themselves. Keep this intuition sharp, and no polynomial will ever intimidate you again.

Animated Solution for Mathematics - Quadratic Equations: If $(x^{2} + p x + 1)$ is a factor of $(a x^{3} + b x + c)$ , then

Select Answer:

Visualized Solution

Polynomial Setup

Finding the Linear Factor

The Polynomial Identity

Expanding the Product

Comparing $x^{2}$ Coefficients

Comparing $x$ Coefficients

Substituting $p$

Final Relation

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Coefficients

Phase 1

Phase 2

Phase 3

Conclusion

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