How to Solve a Binomial Theorem Coefficient Comparison Problem?

Full Video Transcript

Below is the complete transcript from the video explanation:

1. Here’s a classic JEE Main coefficient comparison question. Let’s read it slowly and then we’ll set up equations together. Coefficient of x equals negative 56. This gives our first equation. The x² and x³ coefficients are zero. This will yield two more equations.
2. To organize the work, let’s name the big power as Q(x) and the whole product as P(x). Then we can compare coefficients cleanly. By the binomial theorem, the coefficient of x^k in Q(x) is binomial 26 choose k times negative 2 to the k. We’ll need k equals 0, 1, 2, and 3. Let’s compute those four coefficients carefully.
3. Now use the product rule for coefficients. For P(x) equals the quadratic times Q(x), the coefficient of x^r comes from shifting by degrees 2, 1, and 0. Apply r equals 1, 2, and 3 and set them to the given values: minus 56, 0, and 0. Substitute the numbers from Q(x) to get a simple linear system in a, b, and c.
4. Let’s solve it step by step. From the first equation, b equals minus 56 plus 52c. Plug that into the second to express a in terms of c. Now substitute both a and b into the third equation. We’ll simplify calmly and solve for c.
5. Back substitute to find b and a and then add them. b is minus 56 plus 52 times 3, which is 100. a is 1,404 times 3 minus 2,912, which is 1,300.
6. Finally, compute a plus b plus c. So the value of a plus b plus c is 1403. That matches option C.

Step-by-Step Explanation

Binomial expansion problems like this one are really asking one simple question. Can you extract the right coefficients and set up the right equations? Thinking about it this way makes the maths much more intuitive. Here is how to find the coefficient in a binomial expansion step by step for this exact problem.

Step 1: Define the Two Parts

Before plugging in any numbers, split the expression into two manageable parts. Think of it like separating the problem into what you control (the quadratic) and what the binomial theorem handles (the expansion).

• P(x) = ax² + bx + c <- the quadratic with unknowns
• Q(x) = (1 – 2x)²⁶ <- the binomial expansion part

We want the coefficients of x, x², and x³ in the full product P(x) × Q(x).

Step 2: Extract the Given Conditions

This is where the zero conditions become critical. The problem gives us three pieces of information that will translate directly into three equations.

• Coefficient of x   = -56 <- first equation
• Coefficient of x²  =  0 <- second equation
• Coefficient of x³  =  0 <- third equation

Notice that having two zero conditions alongside one non-zero condition is what makes this system solvable with a unique answer. That is the core structure this binomial question relies on.

Step 3: Apply the Binomial Theorem to Extract the Binomial Coefficient

This is the heart of all binomial coefficient problems. The general term formula gives us the coefficient of xᵏ in Q(x) = (1 – 2x)²⁶ as C(26, k) × (-2)ᵏ

Computing for k = 0, 1, 2, and 3:

• k = 0: C(26, 0) × 1 = 1
• k = 1: C(26, 1) × (-2) = -52
• k = 2: C(26, 2) × 4 = 1300
• k = 3: C(26, 3) × (-8) = -20800

These values come directly from the binomial coefficients $)$, which determine each term in the expansion of (1−2x)^26

Step 4: Set Up the Three Equations Using the Product Rule

This is where the binomial expansion problem gets elegant. Using the product rule for coefficients, each power of x in the full product comes from combining matching degrees from P(x) and Q(x):

• Coefficient of x = -56 → Equation 1:  b(1) + c(-52) = -56
• Coefficient of x² = 0 → Equation 2:  a(1) + b(-52) + c(1300) = 0
• Coefficient of x³ = 0 → Equation 3:  a(-52) + b(1300) + c(-20800) = 0

Step 5: Solve the Linear System

From Equation 1: b = -56 + 52c

Substituting into Equation 2 to express a in terms of c. Then substituting both into Equation 3 and solving c = 3

Back substituting:

• b = -56 + 52 × 3 = 100
• a = 1404 × 3 – 2912 = 1300

Step 6: Final Answer

a + b + c = 1300 + 100 + 3 = 1403

The key insight is that the zero-coefficient conditions reduce the problem with three unknowns to an equation in a single variable c. Once c is determined, the remaining variables can be found by back-substitution.

Frequently Asked Questions