College Algebra

[Ashley189]

Could someone help me out with these two questions. I solve one of them, but want to make sure my answer is right or if not figure out what I did wrong. Thanks in advance.

Factor completely relative to integer.

x^4+6x^2+8

Factor completely, relative to the integers, If the polynomial is prime relative to the integers.

x^2+6xy-40y^2

 

[BigGlenntheHeavy]

\(\displaystyle x^4+6x^2+8 \ = \ (x^2+4)(x^2+2)\)

\(\displaystyle x^2+6xy-40y^2 \ = \ x^2+10xy-4xy-40y^2 \ = \ x(x+10y)-4y(x+10y) \ = \ (x+10y)(x-4y)\)

 

[Denis]

x^2+6xy-40y^2

On those, easier if you remove the y's:
x^2 + 6x - 40
(x + 10)(x - 4)
Now put the y's back in ; kapish?

 

[Ashley189]

\(\displaystyle x^4+6x^2+8 \ = \ (x^2+4)(x^2+2)\)

\(\displaystyle x^2+6xy-40y^2 \ = \ x^2+10xy-4xy-40y^2 \ = \ x(x+10y)-4y(x+10y) \ = \ (x+10y)(x-4y)\)

Thanks! I had the first one right and the second one I need more practice on.

 

[Ashley189]

x^2+6xy-40y^2

On those, easier if you remove the y's:
x^2 + 6x - 40
(x + 10)(x - 4)
Now put the y's back in ; kapish?

Yeah, I get what you mean it's way easier without the y's.