factorise:
a) r^6 - 3r^4
b) x^2 + 5x - 14
c)hence solve the equation x^2 + 5x - 14 = 0
2007-05-14 05:54:15 · 7 answers · asked by Nightmare! 2 in Science & Mathematics ➔ Mathematics
a) r^6 - 3r^4
r^4(r^2 - 3)
b) x^2 + 5x - 14
(x + 7)(x - 2)
c)hence solve the equation x^2 + 5x - 14 = 0
From b) the solutions are x = -7, x = 2.
2007-05-14 06:01:04 · answer #1 · answered by Anonymous · 0⤊ 0⤋
a) r^4(r^2 - 3)
b) (x + 7)(x - 2)
c) (x + 7)(x - 2)= 0
x = -7 or 2
2007-05-14 06:01:51 · answer #2 · answered by lilmaninbigpants 3 · 0⤊ 0⤋
r^4(r^6 - 3)
(x+7) (x-2) x= -7 or x = 2
2007-05-14 06:02:18 · answer #3 · answered by Anonymous · 0⤊ 0⤋
a)
r^4.(r² - 3) = r^(4).(r - √3).(r + √3)
b)
(x + 7).(x - 2)
c)
x = - 7, x = 2
2007-05-14 20:53:06 · answer #4 · answered by Como 7 · 0⤊ 0⤋
a r^4(r^2-3)
b. (x+7)(x-2)
c. multiplying two numbers resulting in zero means that one must be zero
thus 1. what if (x+7) = 0 then x =-7
2 (x-2) = 0; then x = 2
both are your answers
2007-05-14 06:01:59 · answer #5 · answered by Maverick 7 · 0⤊ 0⤋
a) r^4(r^2 - 3)
b) (x + 7)(x - 2)
c) (x + 7)(x - 2) = 0
x + 7 = 0 or x -2 = 0
x = -7 or x = 2
2007-05-14 06:01:15 · answer #6 · answered by Ana 4 · 0⤊ 0⤋
a) r^4(r^2 - 3)
b) (x + 7)(x - 2)
c) (x + 7)(x - 2)= 0
x = -7 or 2
Hope this helps!
2007-05-14 06:25:54 · answer #7 · answered by Anonymous · 0⤊ 0⤋