2006-10-31 05:06:45 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
it factors out to:
(3x+1)(2x+5)
2006-10-31 05:09:41 · answer #1 · answered by Yoni 2 · 0⤊ 0⤋
Polynomial f(x) has a element (x ? p) if and provided that f(p) = 0. 6x² + 17x + 5 = 0 x1,2 = (-17 ± Sqrt(17² - 4*6*5) ) / 12 x1,2 = (-17 ± Sqrt(289 - one hundred twenty) ) / 12 x1,2 = (-17 ± Sqrt(169) ) / 12 x1,2 = (-17 ± thirteen ) / 12 x1 = -5/2 x2 = -a million/3 6x² + 17x + 5 = ( x + 5/2 )( x + a million/3 )
2016-11-26 20:54:40 · answer #2 · answered by ? 4 · 0⤊ 0⤋
6x^2 + 17x + 5
= x^2 + 17/6 + 5/6
= x^2 + 17/6 + (17/12)^2 - (17/12)^2 + 5/6
= (x + 17/12)^2 - 289/144 + 5/6
= (x + 17/12)^2 - 169/144
= (x + 17/12 - 13/12)(x + 17/12 + 13/12)
= (x + 1/3)(x + 5/2)
2006-10-31 05:24:09 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(3x+1) * (2x+5)
start with (x+ ) *(x+ ) to get x^2
then solve for 5 - it must be 1 * 5
now you have (x+1) * (x+5) = x^2 + 6x + 5
now simultaneously solve for 6 (2*3) where these factors will be multiplied by 1 and 5 to get 17 or 2*1+3*5=17
2006-10-31 05:23:30 · answer #4 · answered by Anonymous · 0⤊ 0⤋
6x^2 + 17x + 5
(3x+1)(2x+5)
x=-1/3
x=-2.5
2006-10-31 05:11:29 · answer #5 · answered by yupchagee 7 · 0⤊ 0⤋
6x^2 + 17x +5
=6x^2+2x+15x+5
=2x[3x+1]+5[3x+1]
=[2x+5][3x+1]
2006-10-31 06:58:13 · answer #6 · answered by Anonymous · 0⤊ 0⤋
you have to find 2 integers whose product is 6*5=30 and sum is 17
the integers are 15 and 2
6x^2+17x+5
=6x^2+2x+15x+5
=2x(3x+1)+5(3x+1)
=(2x+5)(3x+1)
2006-10-31 05:20:12 · answer #7 · answered by grandpa 4 · 0⤊ 0⤋
=6x^2+2x+15x+5
=2x[3x+1]+5[3x+1]
=[2x+5][3x+1]
2006-10-31 05:15:14 · answer #8 · answered by openpsychy 6 · 0⤊ 0⤋
Jesus chick, do you're own homework, this is the second math question of yours I have seen in the past 30 sec.
2006-10-31 05:09:27 · answer #9 · answered by Anonymous · 0⤊ 1⤋
(3x+1)*(2x+5)
Doug
2006-10-31 05:10:38 · answer #10 · answered by doug_donaghue 7 · 0⤊ 0⤋