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?
$\begingroup$
$\endgroup$
4
-
4$\begingroup$ Do you think $(1+2)^2=1^2+2^2$? $\endgroup$– DidMar 12, 2012 at 8:47
-
2$\begingroup$ 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: $$ (3\cdot 4+2)^2 = 2^2(3\cdot 2+1)^2 = 4\cdot 7^2 = 196 $$ $\endgroup$– SBFMar 12, 2012 at 8:49
-
1$\begingroup$ Why dont you simplify it first and then open the square, and get (12+2)^2 = 14^2 = 196 $\endgroup$– TomarinatorMar 12, 2012 at 9:34
-
3$\begingroup$ How did this got tagged under number theory I wonder. $\endgroup$– TomarinatorMar 12, 2012 at 11:23
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
2
$(3 \times 4 + 2)^2 = (12 + 2)^2 = 14^2 =196$ 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: $$(x+y)^2 = x^2 + 2 x y +y^2.$$
-
-
1$\begingroup$ 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 $(2s)^2 = 2^2 \times s^2$. $\endgroup$– HenryMar 12, 2012 at 9:04
$\begingroup$
$\endgroup$
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: $$\begin{align} (3 \times 4 + 2)^2 & = & 14^2 \\ & = & 14 \times 14 \\ & = & 196 \end{align} $$
$\begingroup$
$\endgroup$
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:
$$x=(a+b)(a+b)=a\cdot a+2\cdot a\cdot b+a\cdot a = a^2+2ab+b^2$$
$\begingroup$
$\endgroup$
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: $$(ab)^m = a^mb^m$$
answered Sep 21, 2012 at 15:46
$\begingroup$
$\endgroup$
No, you can't distribute the power on any operator, power is distributed on $\times$ and $\div$ not on $+$ and $-$
