factor completely with respect to the integers.
3x^2-48
thanks!
2007-07-09 11:41:28 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
GCF = 3
3(x^2 - 16)
3(x-4)(x+4) diff. of squares
2007-07-09 11:45:19 · answer #1 · answered by gfulton57 4 · 0⤊ 0⤋
Hello,
Factor out a 3 giving us 3(x^2 - 16) so now we have the difference of two square so we have 3(x-4)(x+4)
Hope This Helps!!
2007-07-09 18:45:37 · answer #2 · answered by CipherMan 5 · 0⤊ 0⤋
Take out a common factor of 3
3(x^2-16)
This factorises using the difference of two squares to
3(x+4)(x-4)
2007-07-09 18:47:16 · answer #3 · answered by Anonymous · 0⤊ 0⤋
factor out gcf 3(x^2-16)
factor diff. of sq. 3(x+4)(x-4)
2007-07-09 18:45:31 · answer #4 · answered by JAM 3 · 0⤊ 0⤋
3x^2-48 = 3 * (x^2-16) = 3 * (x^2-4^2) = 3 * (x+4) * (x-4)
2007-07-09 18:47:10 · answer #5 · answered by oregfiu 7 · 0⤊ 0⤋
3(x^2-16)=0
(x-4) (x+4) = 0
x=4, -4
2007-07-09 18:45:47 · answer #6 · answered by :-) 1 · 0⤊ 1⤋