2006-08-26 05:14:28 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
10m^3n^2 - 15m^2n+25m=
5m(2m^2n^2 - 3mn + 5)
substitute in the second factor mn = x
you get
5m(2x^2 -3x +5)
Which doesn't factor using the AC method and has a descrimant, b^2-4ac, of -31 and therefore has no real roots.
So, unless you are studying complex arithematic, you are done factoring at:
5m(2(mn)^2 - 3mn +5)
2006-08-26 05:54:40 · answer #1 · answered by tbolling2 4 · 0⤊ 0⤋
10m^3n^2 - 15m^2n + 25m
5m(2m^2n^2 - 3mn + 5) Factor out the GCF
5m( 2mn - 5)(mn - 1 ) Factor the polynomial
5m[(2mn - 5 )( mn - 1)] final answer
2006-08-26 14:06:33 · answer #2 · answered by grrlgenius5173 2 · 0⤊ 0⤋
Set it equal to zero, and separate the variables.
10m^3n^2-15m^2n+25m=0
5m(2m^2n^2-3mn+5)=0
let mn=x
5x/n(2x^2-3x+5)=0
You can now solve the (2x^2-3x+5) with quadratic formula, (which I'm too lazy to do) then resubsititute your mn for x.
2006-08-26 12:20:24 · answer #3 · answered by Wicked Mickey 4 · 0⤊ 0⤋
Equation is:
10 (m^3)(n^2) - 15(m^2)(n) + 25m
Factor m:
m[10(m^2)(n^2) - 15 mn + 25]
Factor a 5
5m[2(mn)^2 - 3 mn + 5]
Use the quadratic equation to solve the quadratic terms.
2006-08-26 12:23:59 · answer #4 · answered by alrivera_1 4 · 0⤊ 0⤋
10mn^3n^2 - 15m^2n + 25m
5m(2m^2n^2 - 3mn + 5)
2006-08-26 15:48:30 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
10m^3n^2-15m^2n+25m
5m(2m^2n^2-3mn+5)
5m(2m^2n^2+2mn-5mn+5)
5m[(2mn-5) (mn+1)]
2006-08-26 13:06:31 · answer #6 · answered by raj 7 · 0⤊ 0⤋
5m(2mn^2 - 3mn + 5)
2006-08-26 12:20:56 · answer #7 · answered by sassy_91 4 · 0⤊ 0⤋
very carefully.
2006-08-26 13:25:27 · answer #8 · answered by cakeeater0119 5 · 0⤊ 0⤋