factoring

[caughn4]

Can you help me factor this problem? I don't know where to begin.

Do you use the distributive law when factoring?

24x^2 + 14xy +2y^2

 

[galactus]

Can you help me factor this problem? I don't know where to begin.

Do you use the distributive law when factoring?

$\displaystyle 24x^{2} + 14xy +2y^{2}$

Think of the x terms as constants and the y the variable you're factoring. Throw out the x's for now. We'll put them back when we're done.

We see that 2 is a common factor among the coefficients, so factor that out:

$\displaystyle 2(y^{2}+7y+12)$

Now, what two numbers when multiplied equal 12 and when added equal 7?.

How about 3 and 4?.

$\displaystyle 2(3+y)(4+y)$

Now, put the x's back:

$\displaystyle 2(3x+y)(4x+y)$

and that's it.

 

[Deleted member 4993]

Finally check your answer using distributive law.

$\displaystyle 2(3x+y)(4x+y) \ \ = \ \ (6x + 2y)(4x + y) \ \ = \ \ 24x^2 + 6xy + 8xy + 2y^2 \ \ = \ \ 24x^2 + 14xy + 2y^2$

Checks....

 

[caughn4]

Thank you both...I'm drowning in a sea of formulas. They gave me intermediate instead of beginners and its too late to drop it. I havent had algebra in twenty-five yrs. thanks alot

 

[caughn4]

49y^4-81

 

[caughn4]

how to factor completely?
49y^4 - 81

 

[BigGlenntheHeavy]

$\displaystyle 49y^4-81 \ = \ 7^2(y^2)^2 \ - \ 9^2 \ = \ (7y^2)^2-9^2 \ = \ (7y^2+9)(7y^2-9)$

$\displaystyle = \ (\sqrt{7}y+3i)(\sqrt{7}y-3i)(\sqrt{7}y+3)(\sqrt{7}y-3)$

$\displaystyle Therefore, \ 49y^4-81 \ = \ (\sqrt{7}y+3i)(\sqrt{7}y-3i)(\sqrt{7}y+3)(\sqrt{7}y-3)$

$\displaystyle Complete factoring \ of \ complex \ set.$

 

[BigGlenntheHeavy]

$\displaystyle Factor, \ if \ possible, \ 24x^2+14xy+2y^2$

$\displaystyle 1st, \ always \ pull \ out \ common \ factors, \ to \ wit: \ 2(12x^2+7xy+y^2)$

$\displaystyle Next, \ if \ the \ trinomial \ isn't \ prime, \ factor \ it.$

$\displaystyle 2(12x^2+4xy+3xy+y^2), \ grouping$

$\displaystyle 2[4x(3x+y)+y(3x+y)]$

$\displaystyle 2(3x+y)(4x+y) \ and \ you \ are \ done.$

$\displaystyle I'll \ leave \ you \ to \ do \ the \ check. \ Remember, \ factoring \ can \ always \ be \ checked; \ just \ reverse \ the$

$\displaystyle process.$