Question

Let $(1+ax)(1-2x)^{3}=\sum_{n=0}^{4}a_{n}x^{n}$, where $a$ is a constant. If $a_{2}=0$, then the value of $a$ is

• A. 1
• B. 9
• C. 3
• D. 5
• E. 2

Hint:

Logic Tip: When asked for a specific coefficient in a product of polynomials, you don't need to expand the entire expression. Only calculate the specific combinations of terms that multiply to give the desired power of $x$.

Answer 1

Answer: (E)

Concept:
The given equation represents a polynomial expansion. The term $a_2$ refers strictly to the coefficient of the $x^2$ term in the fully expanded expression. We can find this by expanding the terms up to the $x^2$ power and ignoring higher powers.
Step 1: Expand the cubic binomial term.
We only need to expand $(1-2x)^3$ up to the $x^2$ term. Using the binomial expansion $(A-B)^3 = A^3 - 3A^2B + 3AB^2 - B^3$: $$(1-2x)^3 = 1^3 - 3(1)^2(2x) + 3(1)(2x)^2 - (2x)^3$$ $$(1-2x)^3 = 1 - 6x + 12x^2 - 8x^3$$
Step 2: Multiply by the linear term $(1+ax)$.
Now we multiply this result by $(1+ax)$, focusing only on the products that yield $x^2$: $$(1+ax)(1 - 6x + 12x^2 - 8x^3)$$ To get an $x^2$ term, we can multiply: 1. The constant term of the first factor (1) with the $x^2$ term of the second factor ($12x^2$). $\rightarrow 1 \cdot 12x^2 = 12x^2$ 2. The $x$ term of the first factor ($ax$) with the $x$ term of the second factor ($-6x$). $\rightarrow ax \cdot (-6x) = -6ax^2$
Step 3: Combine the $x^2$ terms to find $a_2$.
The total $x^2$ term is $(12 - 6a)x^2$. Therefore, the coefficient is: $$a_2 = 12 - 6a$$
Step 4: Solve for a.
We are given that $a_2 = 0$: $$12 - 6a = 0$$ $$6a = 12$$ $$a = 2$$