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factoring
- Thread starter caughn4
- Start date
caughn4 said: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.
D
Deleted member 4993
Guest
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....
\(\displaystyle 2(3x+y)(4x+y) \ \ = \ \ (6x + 2y)(4x + y) \ \ = \ \ 24x^2 + 6xy + 8xy + 2y^2 \ \ = \ \ 24x^2 + 14xy + 2y^2\)
Checks....
BigGlenntheHeavy
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\(\displaystyle 49y^4-81 \ = \ 7^2(y^2)^2 \ - \ 9^2 \ = \ (7y^2)^2-9^2 \ = \ (7y^2+9)(7y^2-9)\)
\(\displaystyle = \ (\sqrt{7}y+3i)(\sqrt{7}y-3i)(\sqrt{7}y+3)(\sqrt{7}y-3)\)
\(\displaystyle Therefore, \ 49y^4-81 \ = \ (\sqrt{7}y+3i)(\sqrt{7}y-3i)(\sqrt{7}y+3)(\sqrt{7}y-3)\)
\(\displaystyle Complete factoring \ of \ complex \ set.\)
\(\displaystyle = \ (\sqrt{7}y+3i)(\sqrt{7}y-3i)(\sqrt{7}y+3)(\sqrt{7}y-3)\)
\(\displaystyle Therefore, \ 49y^4-81 \ = \ (\sqrt{7}y+3i)(\sqrt{7}y-3i)(\sqrt{7}y+3)(\sqrt{7}y-3)\)
\(\displaystyle Complete factoring \ of \ complex \ set.\)
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Factor, \ if \ possible, \ 24x^2+14xy+2y^2\)
\(\displaystyle 1st, \ always \ pull \ out \ common \ factors, \ to \ wit: \ 2(12x^2+7xy+y^2)\)
\(\displaystyle Next, \ if \ the \ trinomial \ isn't \ prime, \ factor \ it.\)
\(\displaystyle 2(12x^2+4xy+3xy+y^2), \ grouping\)
\(\displaystyle 2[4x(3x+y)+y(3x+y)]\)
\(\displaystyle 2(3x+y)(4x+y) \ and \ you \ are \ done.\)
\(\displaystyle I'll \ leave \ you \ to \ do \ the \ check. \ Remember, \ factoring \ can \ always \ be \ checked; \ just \ reverse \ the\)
\(\displaystyle process.\)
\(\displaystyle 1st, \ always \ pull \ out \ common \ factors, \ to \ wit: \ 2(12x^2+7xy+y^2)\)
\(\displaystyle Next, \ if \ the \ trinomial \ isn't \ prime, \ factor \ it.\)
\(\displaystyle 2(12x^2+4xy+3xy+y^2), \ grouping\)
\(\displaystyle 2[4x(3x+y)+y(3x+y)]\)
\(\displaystyle 2(3x+y)(4x+y) \ and \ you \ are \ done.\)
\(\displaystyle I'll \ leave \ you \ to \ do \ the \ check. \ Remember, \ factoring \ can \ always \ be \ checked; \ just \ reverse \ the\)
\(\displaystyle process.\)