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Algebra Q (1 Viewer)

Green Yoda

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(h+3i)^2 -49j^2

I got (h^2+6hi+9i^2) (j-7)(j+7)

Answer: (h+3i+7j)(h+3i-7j), Can someone explain how did they get this?
 

jathu123

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difference of two squares: a^2 - b^2 = (a-b)(a+b)
in here, a is h+3i and b is 7j.
therefore the answer is (h+3i-7j)(h+3i+7j)
 

drsabz101

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Automatically, you can see that :
1. (h+3i)^2 = (h+3i)(h+3i) - don't bother expanding there is no need

2. then you have - 49j^2 . You take the square root of 49 which is 7. Therefore, this is equal to : - (7j)^2 .
The negative sign is really important.

What happens is you end up with an equation like this:


(h+3i)(h+3i) - (7j) x (7j)

There is a difference of 2 squares here which is how they get :

(h+3i+7j)(h+3j-7j).

Tried to explain it as best as I can
 

BLIT2014

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<a href="https://www.codecogs.com/eqnedit.php?latex=\left(h&plus;3i\right)^2-49j^2&space;=\left(7j&plus;\left(h&plus;3i\right)\right)\left(-7j&plus;\left(h&plus;3i\right)\right)&space;=\left(7j&plus;\left(h&plus;3i\right)\right)\left(-7j&plus;\left(h&plus;3i\right)\right)" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\left(h&plus;3i\right)^2-49j^2&space;=\left(7j&plus;\left(h&plus;3i\right)\right)\left(-7j&plus;\left(h&plus;3i\right)\right)&space;=\left(7j&plus;\left(h&plus;3i\right)\right)\left(-7j&plus;\left(h&plus;3i\right)\right)" title="\left(h+3i\right)^2-49j^2 =\left(7j+\left(h+3i\right)\right)\left(-7j+\left(h+3i\right)\right) =\left(7j+\left(h+3i\right)\right)\left(-7j+\left(h+3i\right)\right)" /></a>
 

Green Yoda

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Ah, i see thank you everyone!
Also what will happen of there are 4+ terms while differencing the square?
 

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