Factor 3x²−16x+20. Rewrite the trinomial with the x-term expanded, using the two factors.

Answer 1

3x^2 -16x + 20 = (3x-10)(x-2)

using FOIL

= 3x^2 --6x -10x + 20

combine the x terms to get

3x^2-16x + 20

factoring is trial and error, but look for integer factors of the constant term: 4x5, 2x10 or 1x20

and integer factors of the x^2 term: 1x3

try to combine them so that (ax-b)(cx-d) where bd =20, ac= 3 and bc+ad = 16

just try a few, there aren't that many combinations. After a try or 2 or 3, or 4 you'll succeed in factoring it Just stumble onto the solution

you might try (3x-20)(x-1) nope, (3x-4)(x-5) nope, (3x-2)(x-10) nope, (3x-10)(x-2) BINGO

-10 and -6 sum to -16 for the x term coefficient. -16x = -10x - 6x

you know you need both integer factors of the constant term to be negative, to get -16x. Only other choice is two positive terms which can never sum to -16x

IF you really can't stumble on to the factors

use the quadratic formula (or complete the square in problems where that's easier)

x = 16/6 + or - (1/6)sqr(16^2 -4(3)(20) = 8/3 + or - (1/6)= 8/3 + or - 4/6 =

8/3 + or - 2/3 = 10/3 or 2

x = 2 or 10

x-10 = 0

x-2 = 0

(x-2)(x-10) are the factors of 3x^2 -16x + 20

x^2 -10x -6x + 20

= (x^2-10x) - (6x-20)

= x(x-10)- 2(3x-10)

= (x-2)(x-10)