i have two problems...
4x^2y^3 + 6xy
and
y^2 - 22y + 72
thank you
2007-06-05 05:52:00 · 6 answers · asked by christianpunkgirl 2 in Science & Mathematics ➔ Mathematics
4x^2y^3 + 6xy
= 2xy(2xy^2 + 3)
y^2 - 22y + 72
= y^2 - 4y - 18y + 72
= y(y - 4) - 18(y - 4)
= (y - 4) (y - 18)
2007-06-05 05:58:12 · answer #1 · answered by raja 3 · 0⤊ 0⤋
4x^2y^3 + 6xy
= 2xy(2xy^2 + 3)
y^2 - 22y + 72 [factorization by grouping]
= y^2 - 4y - 18y + 72
= y(y - 4) - 18(y - 4)
= (y - 4) (y - 18)
2007-06-05 13:12:10 · answer #2 · answered by Krish 5 · 0⤊ 0⤋
4x^2y^3+6xy= 2xy(2xy^2+3)
y^2-22y+72= (y-4)(y-18)=0 y=4 or 18
2007-06-05 13:58:31 · answer #3 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
4x^2y^3 + 6xy =
2xy(2xy^2 + 3)
y^2 - 22y + 72=
(y - 4) x (y - 18)
2007-06-05 12:58:00 · answer #4 · answered by wow_bill 7 · 0⤊ 0⤋
Question 1
4x²y³ + 6xy = 2xy.(2xy² + 3)
Question 2
y² - 22y + 72 = (y - 4).(y - 18)
2007-06-05 14:24:03 · answer #5 · answered by Como 7 · 0⤊ 0⤋
common factor is 2xy
4x^2y^3 + 6xy = 2xy(2xy^2 +3)
+72 is positive, so look ofr factors of it with the same signs.
-22 is negative, so find the two factors (above) that sum to 22 and you know their signs will be negative.
18x4 = 72
18 + 4 = 22
So it factors as (x - 4)(x - 18)
2007-06-05 13:01:46 · answer #6 · answered by jcsuperstar714 4 · 0⤊ 0⤋