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


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.


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:

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

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    $\begingroup$ This makes everything crystal. +1 Absolutely. Now intend to keep foil in the toolkit. $\endgroup$ – usernvk May 9 '13 at 5:58
  • $\begingroup$ Great! Keep it up, looks like you're learning a lot! :) $\endgroup$ – Alex Wertheim May 9 '13 at 6:07
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    $\begingroup$ @AWertheim It is worth mentioning that what's really going on here is the distributive law; that is we have $$(2-4x)(2-4x)=2(2-4x)+(-4x)(2-4x)=2\cdot 2+2\cdot (-4x)+(-4x)\cdot 2+(-4x)\cdot (-4x)$$ $\endgroup$ – Alex Becker May 9 '13 at 6:10
  • $\begingroup$ Right you are @AlexBecker, thanks for being thorough where I wasn't. $\endgroup$ – Alex Wertheim May 9 '13 at 6:20
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    $\begingroup$ @Alex Now that you mention it, disbributivity is in the text. I see the difference now between reading and doing exercises. $\endgroup$ – usernvk May 9 '13 at 7:03

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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,


And so,


As required

  • $\begingroup$ Is this where $(a - b)^2 = a^2 - 2ab + b^2$ came from? +1. Instant visualization. $\endgroup$ – usernvk May 9 '13 at 6:56
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    $\begingroup$ @usernvk Yes, the same logic can also be applied to a cube to give an expanded representation for $(a-b)^3$. $\endgroup$ – Ethan May 9 '13 at 7:00

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

  • $\begingroup$ @manthanomen thanks $\endgroup$ – iostream007 May 9 '13 at 6:00
  • $\begingroup$ Now I see why we spent all those hours years ago memorizing this formula. +1 $\endgroup$ – usernvk May 9 '13 at 6:01

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