(b-a)(a+b)+(c+b-a)(c-b-a)-(b-c)^2simplify true Binomial

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Simplify `(b-a)(a+b)+(c+b-a)(c-b-a)-(b-c)^2` :

(1) `(b-a)(b+a)=b^2-a^2` (The difference of two squares.)

(2) `(c+b-a)(c-b-a)=c^2-bc-ac+bc-b^2-ab-ac+ab+a^2`

`=c^2-2ac-b^2+a^2`  Using the extended distributive property

(3) `(b-c)^2=(b-c)(b-c)=b^2-2bc+c^2`

So putting them all together we get:

`b^2-a^2+c^2-2ac-b^2+a^2-(b^2-2bc+c^2)`

`=b^2-a^2+c^2-2ac-b^2+a^2-b^2+2bc-c^2`

`=-2ac-b^2+2bc`

So the expression simplifies to `-b^2-2ac+2bc`

degeneratecircle's profile pic

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

Just a small typo above: The last line should say `-b^2-2ac+2bc` (see the line directly above that, which is correct).

Since I'm responding, I can add a slightly helpful observation.

`(c+b-a)(c-b-a)=[(c-a)+b][(c-a)-b]=(c-a)^2-b^2,`

which should save a little time.

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