please how can y resolve this? 3x at power 2 -x-2 pls help me?

Question

it's my homework pls need help

Answer 1

Do you mean you need to factor 3x^2 - x - 2? It will be of the form (3x + a)(x + b), because that's the only way you'd get 3x^2. So you need to solve a and b such that ab = -2 and ax + 3bx = -x ==> a + 3b = -1. It looks to me like you can only get ab = -2 if a and b are -2 and 1, or 2 and -1. So by trial and error, the answer is (3x + 2)(x - 1) = 3x^2 - x - 2.

Answer 2

For quadratic equations you can use the quadratic formula.

3x² - x - 2 = 0

x = -b ± √[b² - 4*a*c] / (2a)

x = -(-1) ± √[(-1)² - 4*3*-2] / (2*3)
x = -(-1) ± √[1 + 24] / 6
x = 1 ± √25 / 6
x = 1 ± 5 / 6
x = 6/ 6 (or) -4/ 6
x = 1 (or) -2/ 3

Answer 3

Ok, so we are factoring it, correct?

(-3x-2)(-x+1)

so that when you multiply it out (multiply the firsts, outsides, insides, & lasts) you get
3(x)^2 +2x-3x-2

then combine the +2x & the -3x to make
3(x) ^2 -x- 2

Answer 4

your question is not clear
if
3x^2-x-2=0
then

x1=1
x2=-2/3

if you need to factor the given expresion
3x^2-x-2=3(x-1)(x+2/3)

Answer 5

3x^2-x-2

Just factor it:
(3x+2))(x-1)