What is the coefficient of $x^2$ in the polynomial..

Question

What is coefficient of $x^2$ in polynomial $(1-4x)^6(1+3x)^8$ ? I know how to do it for $(1-4x)^6$ for example, but how to do it for the product of two polynomials?

Answer 1

Strategy:

1. Compute $\alpha, \beta$ in $$(1-4x)^6=1+\alpha x + \beta x^2 +O(x^3)$$

2. Compute $\delta, \gamma$ in $$(1+3x)^8=1+\delta x + \gamma x^2 + O(x^3)$$

3. The answer you seek is $\beta+\gamma+\alpha\delta$.

Answer 2

If you have two polynomials $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ and $b_mx^m\dots b_0$ then the coefficient of $x^2$ in the product is just $b_2a_0+b_1a_1+b_0a_2$. You say that you can find each of these terms.

Answer 3

Basically, you just need to find all of the terms of degree 0, 1, or 2 in each of the two factors, and see what products among them have the desired degree. That's because any higher order terms won't be of interest in the final product.

Answer 4

Let $f(x) = (1-4x)^6(1+3x)^8 = uv$, where $u = (1-4x)^6$ and $v=(1+3x)^8$ \begin{align*} f''(x) &= u'' v + 2 u'v' + v''\\ &= 6 \cdot 5 \cdot (-4)^2 (1-4x)^4 (1+3x)^8 + 2 \cdot 6 \cdot (-4) \cdot 8 \cdot 3 (1-4x)^5 (1+3x)^7 \\ & \qquad + 8 \cdot 7 \cdot (3)^2 (1-4x)^6 (1+3x)^6 \end{align*} Hence coefficient of $x^2$ is $\frac{1}{2!}f''(0) = (480 - 1152 + 504)/2 = -84 $