Factoring all real #s?

Question

Math professor gave me this problem.
3x²-19y²
Attempted to solve it, thinking it was Prime.
(√3x)² - (√19y)²
(√3x - √19y)(√3x + √19y)

My check doesn't work out though. (Using FOIL)
(√3x - √19y)(√3x + √19y)
3x + √57xy - √57xy - 19y
3x - 19y 

Now the professor marked it as correct (I didn't put the check on the paper I turned in) but I would like to know if it is solvable and if so, how to do it. You can copy my symbols if you don't know how to insert them.

Anonymous · Accepted Answer

You did your check wrong. √3x multiplied by itself is 3x², same with - √19y and √19y, it's -19y².
When you check, you do
(√3x-√19y)(√3x+√19y)
3x²+√57xy-√57xy-19y²
3x²-19y²,it checks.
When you square an exponent (i.e., 2x²), you get (√2x)², because √2x times √2x is 2x².
So, your answer is correct.

Mrs. Lovely · Answer

Your check was wrong...you didn't multiply the x's or the y's to get the squares back.

(â3x - â19y)(â3x + â19y)
3xÂ² + â57xy - â57xy - 19yÂ²
3xÂ² - 19yÂ²

? · Answer

Factor: 3x^2 - 19y^2
= (â3x + â19y)(â3x - â19y)
= 3x + â57xy - â57xy + 19y
= 3x + 19y

Anonymous · Answer

(â3x - â19y)(â3x + â19y)

â3x times â3x is not, 3x, it's 3xÂ²
â3 * â3  * x * x =  3xÂ²
same for the 19yÂ²
It works out

? · Answer

You are factoring only.
Your steps are correct.

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Factoring all real #s?

5 Answers