{"id":1661,"date":"2026-03-06T16:11:29","date_gmt":"2026-03-06T16:11:29","guid":{"rendered":"https:\/\/www.think10x.ai\/?p=1661"},"modified":"2026-03-12T12:48:57","modified_gmt":"2026-03-12T12:48:57","slug":"binomial-coefficient-problem","status":"publish","type":"post","link":"https:\/\/www.think10x.ai\/blog\/binomial-coefficient-problem\/","title":{"rendered":"How to Solve a Binomial Theorem Coefficient Comparison Problem?"},"content":{"rendered":"\t\t
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Question<\/h2>

If the coefficient of x in the expansion of (ax\u00b2 + bx + c)(1 – 2x)\u00b2\u2076<\/strong> is -56<\/strong> and the coefficients of x\u00b2<\/strong> and x\u00b3<\/strong> are both zero, then a + b + c<\/strong> is equal to:<\/span><\/p>

(A) 1300 <\/span><\/p>

(B) 1500 <\/span><\/p>

(C) 1403 <\/span><\/p>

(D) 1483<\/span><\/p>

Quick Answer<\/h2>

The correct value of a + b + c<\/strong> is 1403<\/strong>, which corresponds to Option C<\/strong>. This might seem like a lot of steps, but it makes sense once you break the binomial expansion problem into three simple binomial coefficient<\/strong> equations. With the coefficients of x\u00b2<\/strong> and x\u00b3<\/strong> both set to zero, the linear system becomes very clean to solve.<\/span><\/p>

Bottom Line<\/h3>

a + b + c<\/strong> = 1403<\/strong><\/span><\/p>

Even with a 26th power binomial<\/strong>, the zero coefficient<\/strong> conditions mean only about three equations<\/strong> are needed to find a, b, and c individually.<\/span><\/p>

How to Solve This Binomial Coefficient Problem: Video Walkthrough<\/h2>

We created a step-by-step<\/a> video walkthrough of this binomial expansion<\/strong> problem. Watch how we apply the binomial theorem formula from scratch, building each coefficient of Q(x) before combining them into the final answer.<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t

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Full Video Transcript<\/h2>

Below is the complete transcript from the video explanation:<\/span><\/p>

  1. Here’s a classic JEE Main<\/strong> coefficient comparison question. Let’s read it slowly and then we’ll set up equations together. Coefficient<\/strong> of x equals negative 56<\/strong>. This gives our first equation. The x\u00b2<\/strong> and x\u00b3<\/strong><\/span>\u00a0coefficients are zero. This will yield two more equations.<\/span><\/li>
  2. To organize the work, let’s name the big power as <\/span>Q(x)<\/strong> and the whole product as <\/span>P(x)<\/strong>. Then we can compare coefficients cleanly. By the binomial theorem, the coefficient of x^k<\/strong> in <\/span>Q(x)<\/strong> is binomial 26<\/strong> choose k times negative 2<\/strong> to the k. We’ll need k equals 0, 1, 2, and 3. Let’s compute those four coefficients<\/strong> carefully.<\/span><\/li>
  3. Now use the product rule for coefficients. For <\/span>P(x)<\/strong> equals the quadratic times <\/span>Q(x)<\/strong>, the coefficient of x^r<\/strong> 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<\/strong>, 0<\/strong>, and 0<\/strong>. Substitute the numbers from <\/span>Q(x)<\/span> to get a simple linear system in a, b, and c.<\/span><\/li>
  4. Let’s solve it step by step. From the first equation, b equals minus 56<\/strong> plus 52c<\/strong>. Plug that into the second to express a<\/strong> in terms of c<\/strong>. Now substitute both a and b into the third equation. We’ll simplify calmly and solve for c.<\/span><\/li>
  5. Back substitute to find b and a and then add them. b<\/strong> is minus 56 plus 52 times 3<\/strong>, which is 100<\/strong>. a is 1,404<\/strong> times 3<\/strong> minus 2,912<\/strong>, which is 1,300<\/strong>.<\/span><\/li>
  6. Finally, compute a<\/strong> plus b<\/strong> plus c<\/strong>. So the value of a plus b plus c is 1403<\/strong>. That matches option C.<\/span><\/li><\/ol>

    Step-by-Step Explanation<\/h2>

    Binomial expansion problems like this one are really asking one simple question. Can you extract the right coefficients<\/strong> 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<\/strong> step by step for this exact problem.<\/p>

    Step 1: Define the Two Parts<\/h3>

    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).<\/span><\/p>