Rules for multiplication serge lange problem

Question

I have recently started to get interested in math for self learning and recently acquired Serge Langs "Basic Mathematics"

I found myself stuck, trying to figure out one of his proofs regarding more than two factors in multiplication.

The example: Expand the expression (2x + 1)(x - 2)(x + 5)

as a sum of powers of x multiplied by integers

first of we multiply the two factors:

(2x + 1)(x - 2)

= 2x(x - 2) + 1(x - 2)

= 2x^2 - 4x + x - 2

= 2x^2 - 3x - 2

here is my first problem: how come it turns out to be 3x? we have (4x + x)? or is it (-4x + x)?

Second problem comes from the second part where we multiply the the above expression with the last factor:

(2x^2 - 3x - 2)( x + 5)

= (2x^2 - 3x - 2)x + (2x^2 - 3x - 2)5

= 2x^3 - 3x^2 - 2x + 10x^2 - 15x - 10

which in the book turns out to be:

= 2x^3 + 7x^2 - 17x - 10

How the hell did we end up here? I am not following this, could somebody care to explain and i would be very grateful!

Answer 1

first question:

We have $(-4x+x)=(-4+1)x=-3x$, the sign follows the number.

For your second question,

$$2x^3-3x^2-2x+10x^2-15x-10=2x^3+(-3+10)x^2+(-2-15)x-10$$

Answer 2

I have frequently found tabular multiplication of polynomials to be helpful...

Put one factor along the upper border, one term per column. Put the other factor along the left border, one term per row. In the table, put the product of the term at the top of the column with the term at the left of the row. Then read off the results combining like powers. If oyu write your polynomials in in-/de-creasing order, the like powers will generally be in diagonals in the table.

$$ (2x+1)(x-2) : \begin{array}{c|c c} & 2x & +1 \\ \hline x & 2x^2 & x \\ -2 & -4x & -2 \end{array} \text{,} $$ so $$ (2x+1)(x-2) = 2x^2 -3x -2 \text{.}$$ Then $$ (2x^2 -3x -2)(x+5) : \begin{array}{c|c c} & 2x^2 & -3x & -2 \\ \hline x & 2x^3 & -3x^2 & -2x \\ +5 & 10x^2 & -15x & -10 \\ \end{array} $$ gives us $$ (2x^2 -3x -2)(x+5) = 2 x^3 + 7 x^2 - 17 x -10 \text{.}$$

Notice that we calculate all the same pairs of "one term from this polynomial times one term from the other polynomial" that we calculate by the distribution you are demonstrating. We just record these products in the table and, as we read out the result, we combine the terms with the same power of $x$.