How do I reduce the following expression to its lowest terms..
6x^2 - 96 OVER x^2 - 9x + 20
Thanks!
2007-05-02 19:03:38 · 5 answers · asked by shauna s 1 in Science & Mathematics ➔ Mathematics
6x^2 - 96 OVER x^2 - 9x + 20
= [6(x^2 - 16)] / [(x-4)(x-5)]
= [6(x^2 - 4^2)] / [(x-4)(x-5)]
= [6(x - 4)(x+4)] / [(x-4)(x-5)]
= [6(x+4)] / (x-5)
2007-05-02 19:10:50 · answer #1 · answered by QiQi 3 · 0⤊ 0⤋
6x^2 - 96/ x^2 - 9x + 20
Factor the completely the numerator and denominator
6(x^2 - 16) / (x - 5) (x - 4) =
6( x + 4) (x - 4)/ (x-5)(x-4) =
(x-4) will cancell out
6(x+4)/(x-5) ==> the answer
2007-05-02 20:05:41 · answer #2 · answered by detektibgapo 5 · 0⤊ 0⤋
6x^2 - 96 OVER x^2 - 9x + 20
= 6(x^2 - 16) over (x -4)(x-5)
= 6(x-4)(x+4) over (x-4)(x-5)
= 6(x+4) over (x-5)
2007-05-02 19:12:01 · answer #3 · answered by looikk 4 · 0⤊ 0⤋
= 6(x² - 16) / (x - 5).(x - 4)
= 6.(x - 4).(x + 4) / (x - 5).(x - 4)
= 6.(x + 4) / (x - 5)
2007-05-02 19:11:53 · answer #4 · answered by Como 7 · 0⤊ 0⤋
[6(x^2-16)]/[x^2-5x-4x+20]
=[6(x+4)(x-4)]/[x(x-5)-4(x-5)]
=[6(x+4)(x-4)]/[(x-5)(x-4)]
=[6(x+4)]/(x-5)
2007-05-02 19:11:44 · answer #5 · answered by Maths Rocks 4 · 0⤊ 0⤋