Basic Mathematics. Trouble with powers and polynomials.

Question

Greets,

I'm hoping to complete the exercises in chapter 1.3 Basic Mathematics, Serge Lang.

The section question is:

Expand the following expressions as sums of powers of $\;x\;$ multiplied by integers. These are in fact called polynomials. You might want to read, or at least look at, the section on polynomials later in the book (Chapter 13, §2).

Chapter 13.2 is of no help and I don't understand what the question is asking.

The answer to one expression:

4) $\qquad(2 - 4x)^2$ $\quad$ is $\quad$ $16x^2 - 16x + 4$

I'm lost here. How does it get there?

EDIT: for learning, @Ethan's answer is the most comprehensive, for study/practice, @AWertheim's answer is invaluable (comments also point to distributivity, essential here), chose the answer that most directly answered this question.

Goes upvotes other good answers by these users

Answer 1

Given: $(2 - 4x)^2$

$=(2-4x)(2-4x)$ Expand.

$= 2(2-4x) -4x(2-4x)$ Rewrite the equation into two terms.

$=4-8x-8x+16x^2$ Gather like terms.

$=4-16x+16x^2$ Answer.

$=16x^2-16x+4$ Rewrite by degree.

Answer 2

Exponentiation means the same thing for polynomials as it does for real numbers. Hence, $(2-4x)^{2} = (2 - 4x)(2 - 4x)$.

To expand this out, we use something called the "FOIL" method, which is just another way of explaining how to apply the property of distributivity. "FOIL" stands for "first, outer, inner, last" - which is another way of saying, "sum the product of the first terms in each expression, the product of the outer terms, the product of the inner terms, and the product of the last terms." Thus, we see:

$(2-4x)(2-4x) = (2)(2) + (-4x)(2) + (-4x)(2) + (-4x)(-4x) = 4 - 8x - 8x + 16*x^{2} = 4 - 16x + 16x^{2}$.

Answer 3

I think your question is expand $(2-4x)^2$. use this formula $(a-b)^2$= $a^2$ + $b^2$ -2*a*b. similarly $(2-4x)^2$ = $2^2$ + $(4x)^2$ - 2 * 2* 4x

= 4+$16x^2$ -16x simple

Answer 4

Sense the area of the entire square is $2*2=4$, and the area of the square is equal to the sum of the smaller areas of each component of the square, we have that $$4=(2-4x)^2+4x(2-4x)+4x(2-4x)+(4x)^2$$ $$=(2-4x)^2+8x(2-4x)+16x^2$$ $$=(2-4x)^2+16x-32x^2+16x^2$$ $$=(2-4x)^2+16x-16x^2$$

So that we have,

$$4=(2-4x)^2+16x-16x^2$$

And so,

$$4-16x+16x^2=(2-4x)^2$$

As required