1. p^4+5p^2+6
2.15r^2-16rt+4t^2
2007-01-30 17:03:45 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1. factor as if it said x² + 5x + 6, you get (x+2)(x+3), then since you replaced p² with x, put it back -- you get (p² + 2)(p² + 3)
2. you need factors of 15•4 that add up to 16, and that's 10 and 6, so
15r² - 16rt + 4t² =
15r² - 10rt - 6rt + 4t² =
5r(3r - 2t) - 6t(3r - 2t) =
(5r - 6t)(3r - 2t)
2007-01-30 17:21:03 · answer #1 · answered by Philo 7 · 1⤊ 0⤋
1. p^4+5p^2+6
(p^2+3)(p^2+2) no real factors of either binomial
2.15r^2-16rt+4t^2
(5r-2t)(3r-2t)
2007-01-31 01:41:13 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋
p^4 + 5p^2 + 6 = (p^2 + 3)(p^2 + 2)
15r^2 - 16rt + 4t^2 = (5t - 2t)(3t - 2t)
2007-01-31 01:21:46 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
p^4+5p^2+6 = p^4 +3p^2 + 2p^2 +6
= p^2(p^2+3) + 2(p^2+3)
= (p^2+3) (p^2+2) ----------------------------------------------Answer 1
15r^2-16rt+4t^2 = 15r^2 -10rt - 6rt + 4t^2
= 5r(3r-2t) - 2t(3r-2t)
= (3r-2t) (5r-2t) ------------------------------------------------Answer 2
2007-01-31 01:26:35 · answer #4 · answered by Mritunjay 2 · 1⤊ 0⤋
1. (p^2+3)(p^2+2)
2. (5r-2t)(3r-2t)
2007-01-31 01:14:53 · answer #5 · answered by Jinkus 2 · 0⤊ 0⤋
(p^2+3)(p^2+2)
(5r-2t)(3r-2t)
2007-01-31 01:15:31 · answer #6 · answered by Sammy Baby 1 · 0⤊ 0⤋