Reverse distributive property simplification

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October 17th 2012, 04:10 PM
xXplosionZz

Reverse distributive property simplification

Ok, so i have a generally clear idea on what then distributive property is. I understand that if a number is outside of a pair of brackets then you multiply that number by all the entities in the brackets although I have no idea how to solve this question. Expand, using the distributive property. Simplify. (5 - 4 times the square root of 3)(-2 + the square root of 3)

(I am only grade 10, so please try to avoid using any complicated (higher grade) math short cuts without clearly describing them) The answer is -22 + 13 square root of 3.

I would greatly appreciate any answer displaying the steps on how to solve this, and what you did each step.(Happy)
October 17th 2012, 05:42 PM
johnsomeone

Re: Reverse distributive property simplification

$(5 - 4 \sqrt{3})(-2 + \sqrt{3})$

$= (5 - 4 \sqrt{3})(-2) + (5 - 4 \sqrt{3})(\sqrt{3})$

$\text{ (That's using the distributive law: } (x)(-2 + \sqrt{3}) = (x)(-2) + (x)(\sqrt{3}).\text{ )}$

$= [ \ (5)(-2) + (- 4 \sqrt{3})(-2) \ ] + [ \ (5)(\sqrt{3}) + (- 4 \sqrt{3})(\sqrt{3}) \ ]$

$\text{ (That's using the distributive law twice again - once for each bracket [ ]. )}$

$= (-10) + (- 4)(-2)\sqrt{3} + (5 \sqrt{3}) + (- 4)(\sqrt{3})^2$

$= -10 + 8\sqrt{3} + 5\sqrt{3} - (4)(3)$

$= -10 + (8+5)(\sqrt{3}) - 12$

$\text{ (That's using the distributive law again: } 8x + 5x = (8 + 5)x \text{. )}$

$= -22 + (13)(\sqrt{3})$

$= -22 + 13 \sqrt{3}$
October 17th 2012, 05:53 PM
xXplosionZz

Re: Reverse distributive property simplification

$= (5 - 4 \sqrt{3})(-2) + (5 - 4 \sqrt{3})(\sqrt{3})$

$\text{ (That's using the distributive law: } (x)(-2 + \sqrt{3}) = (x)(-2) + (x)(\sqrt{3}).\text{ )}$

I don't really understand this first part? how did you pull a -2 out of the first pair of parenthesis and a sqrt{3} out of the second pair?
October 17th 2012, 06:10 PM
xXplosionZz

Re: Reverse distributive property simplification

I apologize as after a decently long pondering on how you have done that I realized you applied the brackets to each other, Would an easy was of saying this law be (x)(a+b) = (ax + bx) ? Thank you so much for you time and effort. :)

My brain sort of throws basic thought out of the window when so many applications of one law start to intermingle. Do you have any advice on how to to view these problems with greater ease in, say, a test situation?
October 17th 2012, 06:58 PM
pickslides

Re: Reverse distributive property simplification

$(a+b)(c+d)= ac+ad+bc+bd$
October 18th 2012, 06:46 AM
HallsofIvy

Re: Reverse distributive property simplification

Quote:

Originally Posted by xXplosionZz

I apologize as after a decently long pondering on how you have done that I realized you applied the brackets to each other, Would an easy was of saying this law be (x)(a+b) = (ax + bx) ? Thank you so much for you time and effort. :)

Yes, that is precisely the "distributive law" which you said you understood.

Perhaps you learned the distributive law as (a+ b)c= ac+ bc. Because multiplication of numbers is commutative, the left side is the same as c(a+ b) and the right side is the same as ca+ cb. So c(a+ b)= ca+ cb.

Was that what you meant by "reverse" distributive property? I had thought you mean going from ab+ ac to a(b+ c)!

My brain sort of throws basic thought out of the window when so many applications of one law start to intermingle. Do you have any advice on how to to view these problems with greater ease in, say, a test situation?[/QUOTE]
October 18th 2012, 02:49 PM
xXplosionZz

Re: Reverse distributive property simplification

Ahhh yes just today I learned that We can use the Foil law. I apologize as I had convinced myself the only way to somehow perform the second step, was to factor. Thank you guys for clearly describing everything :)