Use grouping to factor the polynomial completely
2007-08-01 18:00:59 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^3 + 3x^2 + ax + 3a
x^2(x + 3) + a(x + 3)
(x^2 + a)(x + 3)
2007-08-01 18:03:19 · answer #1 · answered by Anonymous · 0⤊ 0⤋
X^3+3x^2+ax+3a
x^2(x+3)+a(x+3)
(x^2+a)(x+3)
2007-08-01 18:05:42 · answer #2 · answered by Meng-Tzu 4 · 0⤊ 0⤋
x^3 + ax + 3a +3x^2
= x (x^2 + a) + 3 (a + x^2)
= (x + 3) (x^2 + a)
2007-08-01 18:05:38 · answer #3 · answered by vlee1415 5 · 0⤊ 0⤋
Rearrange: x^3 + 3x^2 + ax + 3a
Group: (x^3 +3x^2) + (ax +3a)
Factor: x^2 * (x + 3) + a*(x + 3)
Common factor: (x^2 +a)(x + 3)
FOIL: x^3 + 3x^2 +ax +3a
2007-08-01 18:13:26 · answer #4 · answered by fjblume2000 2 · 0⤊ 0⤋
x^3+ax+3a+3x^2
arrange to be x^3+3x^2+ax+3a
x^2(x+3)+a(x+3)
(x+3)(x^2+a)
2007-08-01 18:10:32 · answer #5 · answered by mathavan velayutham 2 · 0⤊ 0⤋
x^3 + 3x^2 + ax + 3a
x^2(x+3) + a(x+3)
(x^2+a)(x+3)
2007-08-01 18:05:27 · answer #6 · answered by bourqueno77 4 · 0⤊ 0⤋
(x^2+a)(x+3)=x^3+3x^2+ax+3a.
2007-08-01 18:04:33 · answer #7 · answered by David Z 3 · 0⤊ 0⤋
(x^2)(x+3)+a(x+3)=
(x+3)(x^2+a)
2007-08-01 18:07:13 · answer #8 · answered by Anonymous · 0⤊ 0⤋
x(x^2 + a) + 3(x^2 + a) =
(x + 3)(x^2 + a) =
(x + 3)(x + ja)(x - ja)
(j = √-1)
2007-08-01 18:07:29 · answer #9 · answered by Helmut 7 · 0⤊ 0⤋
http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/partial.html
2007-08-01 18:03:16 · answer #10 · answered by Anonymous · 0⤊ 1⤋