I have a question about the bonomial theorem, and in specifically, a question that I want help on. I have worked out the answer, but by manually expanding each and every alternative. However, I thought to myself that there must be a faster way of coming to the answer. I've spent an hour or so thinking about it, but haven't thought of anything. Is there any faster method of coming to the answer of the following question? If so, please explain to me how

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  • $\begingroup$ It is not equal to C either!! When $x=3/2$, $(2x-3)^4=0$ but C is undefined. $\endgroup$ Commented Nov 22, 2011 at 10:09
  • $\begingroup$ An unmentioned test for deciding between D and E is numerical substitution for $x$. Good test values are $0$, $1$, or $-1$ (because they have easy powers!). With $x=0$, we have $(2x-3)^4=(-3)^4=81$; D and E both give $81$, so they both remain candidates. With $x=1$, we have $(2x-3)^4=(-1)^4=1$; here, D gives 55, so it's definitely not equal to the original, and since we know that there's only one answer, we're done! (BTW, E gives 1.) NOTE: The test only tells you when things are NOT alike! Getting equal results (as when we got 81 from D) tells you nothing about whether expressions are equal. $\endgroup$
    – Blue
    Commented Nov 22, 2011 at 11:57
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    $\begingroup$ @Jp McCarthy: No, as rational functions $\rm\ z^6/z^2 = z^4\ $ holds true at $\rm\ z = 0\:.$ $\endgroup$ Commented Dec 25, 2011 at 0:32
  • $\begingroup$ @Bill: I don't see what you are getting at... $\endgroup$ Commented Jan 3, 2012 at 10:56
  • $\begingroup$ @JpMcCarthy The numerator is $0$ as well as the denominator. Usually, when talking about rational functions we say that the value of the functions at points where the numerator and denominator are $0$ is the limit of their value at $x$ as $x$ approaches the point. By this convention, we get that C is $0$ at $x=3/2$. $\endgroup$ Commented Jul 7, 2012 at 0:03

5 Answers 5


It's pretty clear that A, B, and C are equal to $(2x-3)^4$. Therefore, only D and E are left. To see which one is equal to $(2x-3)^4$, I look at the term $x^3$ of $(2x-3)^4$., which is equal to $${4\choose 1}(2x)^3(-3)=-96x^3.$$ Therefore, E is equal to $(2x-3)^4$, and D is not equal to $(2x-3)^4$.

  • $\begingroup$ Sorry, can you please explain how A, B and C are also equal to it? I've just come out of high school in a poor district with sub-par math teaching and I'm trying to catch back up again. $\endgroup$
    – tina nyaa
    Commented Nov 22, 2011 at 9:43
  • $\begingroup$ We know that $(-t)^2=(-t)(-t)=t^2$. Similarly, $(-t)^{2n}=\big((-t)^2\big)^n=t^{2n}$ where $n$ is any positive integer. Therefore, if we take $t=2x-3$ and $n=2$, we have $(3-2x)^4=(-t)^4=t^4=(2x-3)^4$. This proves A. Similarly, the denominator in C is equal to $(3-2x)^2=(2x-3)^2$, which implies that $\frac{(2x-3)^6}{(3-2x)^2}=\frac{(2x-3)^6}{(2x-3)^2}=(2x-3)^{6-2}=(2x-3)^{4}$. For B, we have $(2x-3)(2x-3)^3=(2x-3)^{1+3}=(2x-3)^4$. $\endgroup$
    – Paul
    Commented Nov 22, 2011 at 9:56

You just need a few algebraic properties, perhaps you remember that multiplying two negative numbers produces a positive number $(-2)\cdot(-3)=6$. When you multiply this with another negative number the result becomes negative again. $(-2)\cdot(-3)\cdot(-5)=6\cdot(-5)=-30$. If you look at multiplications with even more negative factors you can see a pattern emerge, that whenever the number of negative factors is odd the result will be negative and positive when the number of negative factors is even.

Then you need to recall, that the power notation $a^n$ is just an abrreviation of a long product $a^n=\underbrace{a\cdot a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}$. Perhaps you also know that when you raise a product to a power you can distribute the power, like so $(a\cdot b)^n=a^ n\cdot b^n$.

The last bit you need to answer the question is, that you always can factor out any term you might need. For example in the expression $(3-2x)$ there is no number $5$ in it. But if I want to have a $5$, or any other number in it, I just have to use $\frac{5}{5}=1$. Because multiplying with $1$ doesn't change anything. So I can put in almost any place I like $$(3-2x)=\left(\frac{5}{5}3-\frac{5}{5}2x\right)=5\left(\frac{3}{5}-\frac{2}{5}x\right)$$ and then factor it out.

If you combine the above it helps to answer your question. First A, you just have to factor out a $(-1)$ $$(3-2x)^4=\bigl((-1)\cdot (-3+2x)\bigr)^4=(-1)^4\cdot (-3+2x)^4=(2x-3)^4$$ then use the properties of the power notation and then you have to realise that $(-1)^4=1$ and multiplying with $1$ doesn't change things, because the $1$ is the identity or neutral element with respect to multiplication. Finally you rearrange $-3+2x=2x-3$, this is possible because addition and with it subtraction is commutative $a+b=b+a$.

This works for C too $$ \frac{(2x-3)^6}{(3-2x)^2}=\frac{(2x-3)^6}{\bigl((-1)\cdot(-3+2x)\bigr)^2}=\frac{(2x-3)^6}{(-1)^2\cdot(-3+2x)^2}=\frac{(2x-3)^6}{(-3+2x)^2}=\frac{(2x-3)^6}{(2x-3)^2}=(2x-3)^4 $$ The last step is just cancellation as in any other fraction. When you do this type of algebraic manipulation more often you will quickly learn these rules by heart. And then realize more general and powerful patterns. For example this swapping $$(a-b)^n=(b-a)^n$$ we did, works whenever $n$ is even, because we factor out $(-1)^n=1$ and a product with an even number of negative factors is always positive.

As for B it's just an application of the abbreviation rule $a^n=\underbrace{a\cdot a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}$ which can also be written recursively as $$a^n=a^{n-1}\cdot a\qquad a^1=a$$ Perhaps you can spot the pattern $$a^4=a^3\cdot a\qquad$$ as $a$ is just a variable, a general placeholder, we have $$(2x-3)^3(2x-3)=(2x-3)^4$$

These algebaric manipulations take practice, it's like learning a language or swimming or playing the piano, it takes practice to spot the patterns. For someone more practised the derivations above are in painfully slow motion.

  • $\begingroup$ Ah! too slow again, why doesn't it show the other answers being written while one works on ones own... ;-) $\endgroup$
    – uli
    Commented Nov 22, 2011 at 10:57
  • $\begingroup$ Slow(er), but amazing answer! Thanks ^^ $\endgroup$
    – tina nyaa
    Commented Nov 23, 2011 at 8:10

It's not $A$, since $(3-2x)^4= ((-1)(2x-3))^4=(-1)^4(2x-3)^4=(2x-3)^4$. (Switching the order of subtraction amounts to factoring out $-1$. Taking an even power of $-1$ makes it $1$.)

It's not $B$, using the law $ a^na^m=a^{n+m}$ :


It's not C), using the law $ {a^n\over a^m}=a^{n-m}$ (and the parenthetical remark above):
$${(2x-3)^6\over (3-2x)^2}= {(2x-3)^6\over (-1)^2 (2x-3)^2} ={(2x-3)^6\over (2x-3)^2} =(2x-3)^{6-2}=(2x-3)^4 $$

So, the answer is D) or E).

Now the "$x$ term" of $(2x-3)^4=$ $$(2x-3)(2x-3)(2x-3)(2x-3)$$ is obtained by picking a $2x$ from one factor and a $-3$ from the others. There are $4 $ ways of doing this, so the $x$ term is $4\cdot 2x\cdot(-3)\cdot(-3)\cdot(-3)=-216x$.

So the answer is D).


Option $A$ is correct because $(a + b)^n = (b + a)^n$

Option $B$ and Option $C$ is correct due to the rule of indices, $\space a^m \times a^n = a^{m+n}$ and $\frac{a^m}{a^n}= a^{m-n}$

we now have to analyze option D and E, Lets try to think using binomial theorem, the $t-th$ term of the expansion $(x+a)^n$ is given by $$x^t \binom{n}{t} a^{n-t}$$

If you look at the options you will see that only the first term and the last term are same the middle one's are all disparate, so we need to check any to spot the faulty one.

Lets check the $4$-th term i.e the coefficient of $x$, using the previously defined formula on the expansion of $(2x-3)^4$ we get, $$ (2x) \binom{4}{1} (-3)^3 = -216x$$

Clearly option $E$ is correct, so your correct answer is option $D$.

  • $\begingroup$ Thanks for your clear answer! I understand the latter part, but can you please explain how A, B and C are also equal to it? I've just come out of high school in a poor district with sub-par math teaching and I'm trying to catch back up again. $\endgroup$
    – tina nyaa
    Commented Nov 22, 2011 at 9:48
  • $\begingroup$ @tina nyaa:I added the corresponding explanation. Is it ok? $\endgroup$
    – Quixotic
    Commented Nov 22, 2011 at 9:54
  • $\begingroup$ Your very first statement is incorrect: $(a-b)^n \ne (b-a)^n$ if $n$ is odd. $\endgroup$
    – Srivatsan
    Commented Nov 22, 2011 at 10:01
  • $\begingroup$ If n is odd we need to multiply by (-1), anyways thanks for pointing that out :), I will leave out the minus part (here) to avoid the obvious complications. $\endgroup$
    – Quixotic
    Commented Nov 22, 2011 at 10:09

Here is the quick trick: at $\rm\ x = 3/2\ $ we have $\rm\ D\ =\ 81 - 81 + 81 - 81 + 81\ =\ 81 \:\ne\: 0\quad$ QED

Probably the polynomial in D was devised so it has this easy evaluation at $\rm\ x = 3/2\:,\:$ viz.

$$\rm\ y\ :=\ \dfrac{2\:x}3\ \Rightarrow\ \frac{D}{81}\ =\ \frac{y^5+1}{y+1}\ =\ y^4 - y^3 + y^2 - y + 1\ \ (= 1\ \ for\ \ y\: =\: 1\:\:\leftrightarrow\ x\: =\: 3/2)$$

But to solve the problem quickly you don't need to discover this structure. Instead, as designed, brute-force prevails: evaluate them all at $\rm\ x = 3/2\ $ to find which lacks this root. This requires only about $10$ seconds of easy mental arithmetic.


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