Chapter 9.5, Problem 38E

To determine

To expand: the expression ${\left(4x-1\right)}^3-2{\left(4x-1\right)}^4$ by using Pascal’s Triangle.

Answer to Problem 38E

  $-512x^4+576x^3-240x^2+44x-3$

Explanation of Solution

Concept Used:

Pascal’s triangle:

The n^th row of Pascal’s triangle gives the coefficients of ${\left(x+y\right)}^n$ .

Consider the given expression ${\left(4x-1\right)}^3-2{\left(4x-1\right)}^4$ .

First use fourth row of Pascal’s triangle to evaluate ${\left(4x-1\right)}^4$ .

The binomial coefficients from the fourth row of Pascal’s Triangle are 1, 4, 6, 4, 1.

(Since here the binomial is representing difference rather than sum, so the signs will be alternate)

So, the expansion is:

  $\begin{array}{c}{\left(4x-1\right)}^4=\left(1\right){\left(4x\right)}^4-4{\left(4x\right)}^3\left(1\right)+6{\left(4x\right)}^2{\left(1\right)}^2-4\left(4x\right){\left(1\right)}^3+\left(1\right){\left(1\right)}^4\\ =256x^4-256x^3+86x^2-16x+1\end{array}$

Now, use third row of Pascal’s triangle to evaluate ${\left(4x-1\right)}^3$ .

The binomial coefficients from the third row of Pascal’s Triangle are 1, 3, 3, 1.

(Since here the binomial is representing difference rather than sum, so the signs will be alternate)

So, the expansion is:

  $\begin{array}{c}{\left(4x-1\right)}^3=\left(1\right){\left(4x\right)}^3-\left(3\right){\left(4x\right)}^2{\left(1\right)}^2+3\left(4x\right){\left(1\right)}^2+\left(1\right){\left(1\right)}^3\\ =64x^3-48x^2+12x-1\end{array}$

Thus,

  $\begin{array}{c}{\left(4x-1\right)}^3-2{\left(4x-1\right)}^4=64x^3-48x^2+12x-1-2\left(256x^4-256x^3+86x^2-16x+1\right)\\ =64x^3-48x^2+12x-1-512x^4+512x^3-192x^2+32x-2\\ =-512x^4+576x^3-240x^2+44x-3\end{array}$