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If I have this $(3 \times 4 + 2)^2$,

How can I simplify it with out the final result.

Do I distribute the $^2$ over each number like this:

$(3^2 \times 4^2 + 2^2)$?

What is the rule?

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    $\begingroup$ Do you think $(1+2)^2=1^2+2^2$? $\endgroup$
    – Did
    Mar 12, 2012 at 8:47
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    $\begingroup$ The rule is $(a+b)^2 = a^2+2ab+b^2$ - but it is not clear, what do you mean by a simplification. You may consider to take $2$ out of brackets: $$ (3\cdot 4+2)^2 = 2^2(3\cdot 2+1)^2 = 4\cdot 7^2 = 196 $$ $\endgroup$
    – Ilya
    Mar 12, 2012 at 8:49
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    $\begingroup$ Why dont you simplify it first and then open the square, and get (12+2)^2 = 14^2 = 196 $\endgroup$ Mar 12, 2012 at 9:34
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    $\begingroup$ How did this got tagged under number theory I wonder. $\endgroup$ Mar 12, 2012 at 11:23

5 Answers 5

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$(3 \times 4 + 2)^2 = (12 + 2)^2 = 14^2 =196$ while $(3^2 \times 4^2 + 2^2) = 9 \times 16 +4 = 144+4 = 148$, so that does not work.

If you want a rule for squares of sums, try: $$(x+y)^2 = x^2 + 2 x y +y^2.$$

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  • $\begingroup$ I want a rule for $(2s + 1)^2$, too $\endgroup$
    – Franceia
    Mar 12, 2012 at 8:58
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    $\begingroup$ From $(x+y)^2 = x^2 + 2 x y +y^2$, let $x=2s$ and $y=1$ to get $(2s)^2+2\times 2s \times 1 + (1)^2$ which you can easily tidy up. Note that $(2s)^2 = 2^2 \times s^2$. $\endgroup$
    – Henry
    Mar 12, 2012 at 9:04
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Always go basic by using the order of operations:

  • Parentheses.
  • Exponents.
  • Multiplication and division.(left to right)
  • Subtraction and addition.(no same order)

Using the PEMDAS rule, first simplify the parentheses, then simplify the exponent(s). We have,$$ 3 \times 4 + 2$$in the parentheses. Notice that again, PEMDAS is applied. Multiplication is done before addition. So, the simplification of the parentheses is as follows. $$\begin{align}3 \times 4 + 2 & = \color{maroon}{3 \times 4} + 2 \\ & = 12 + 2 \\ & = 14 \end{align}$$Now, the exponent. We'd have everything simplified as shown below: $$\begin{align} (3 \times 4 + 2)^2 & = & 14^2 \\ & = & 14 \times 14 \\ & = & 196 \end{align} $$

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No you can't distribute the powers like what you did.

There is a theorem called the binomial theorem that controls this type of operations.

The result is clearly $196$. Here is one way to get this result (steps are simplified so that you can follow)

You can do this to simplify the expression:

$$x=(3\cdot4+2)^2$$

then

$$x=(12+2)(12+2)$$

$$x=(12\cdot12)+2\cdot(12\cdot2)+(2\cdot2)$$

$$x=144+48+4 = 196$$

An expression like:

$$x=(a+b)^2$$

can be written as:

$$x=(a+b)(a+b)=a\cdot a+2\cdot a\cdot b+a\cdot a = a^2+2ab+b^2$$

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No, you can't distribute the power on any operator, power is distributed on $\times$ and $\div$ not on $+$ and $-$

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Can you simplify $$(3\times4 + 2)?$$

then$$(3\times4 + 2)^2$$

by definition $a^2 = a\times a$
let a=$(3\times4 + 2)$

$$ = (3\times4 + 2)\times(3\times4 + 2)$$ multiplication before addition within the parentheses
$$ =(12 + 2) \times(12 + 2)$$ $$ =(14)\times(14)$$ $$= 196$$

The distributive property of multiplication applies to coefficients not to exponents.

a(b+c) = ab+ac

If the 2 was in front as a coefficient then you could write:

$2(3\times 4 +2)$

$= 2\times3\times 4 +2\times 2$

The exponent sort of "distributes" when using the rule of exponents for a power of a product: $$(ab)^m = a^mb^m$$

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