Factor the following polynomials.
2007-08-28 11:26:04 · 8 answers · asked by kellyann 1 in Science & Mathematics ➔ Mathematics
3x^2-33x-36
=3(x^2-11x-12)
=3(x^2+x-12x-12)
=3{x(x+1)-12(x+1)}
=3(x+1)(x-12)
2007-08-28 11:32:29 · answer #1 · answered by Anonymous · 1⤊ 0⤋
3x2/3 - 33x/3 - 36/3=0/3
x^2-11x-12=0
(x-12)(x+1)=0
x-12=0->x=12
x+1=0->x=-1
2007-08-28 11:38:23 · answer #2 · answered by cool j 2 · 0⤊ 0⤋
3x^2-33x-36=0
divide by 3;
x^2-11x-12=0
(x-12)(x+1)=0
x=12,-1 ANS.
2007-08-28 11:38:21 · answer #3 · answered by Anonymous · 0⤊ 0⤋
3x2 - 33x - 36?
= 3(x^2-11x-12)
= 3(x-12)(x+1)
2007-08-28 11:33:28 · answer #4 · answered by ironduke8159 7 · 1⤊ 0⤋
Start by extracting 3 to get 3(x^2 -11x -12). We want two numbers whose product is 12 and whose difference is 11; 1 and 12 qualify. This gives 3(x+1)(x-12).
2007-08-28 11:32:33 · answer #5 · answered by Anonymous · 1⤊ 0⤋
3(x+12)(x-1)
2007-08-28 11:31:07 · answer #6 · answered by carla c 1 · 0⤊ 2⤋
3x^2-33x-36
=3(x^2-11x-12)
=3(x^2+x-12x-12)
=3{x(x+1)-12(x+1)}
=3(x+1)(x-12)
2007-08-28 12:15:38 · answer #7 · answered by John 2 · 0⤊ 0⤋
3(x2-11x-12)
3(x-12)(x+1)
2007-08-28 11:32:38 · answer #8 · answered by D J 3 · 1⤊ 0⤋