Find all the rational zeros of the function?

Question

f(x)= X^3-2X^2 - X +2

i did the plugging and adding by first finding the possible numbers that would make the function equal zero, i plugged in pos. and neg 1 and pos. and neg 2 but i still didn't get zero... help me there is a solution also.

Answer 1

f(x) = x^3 - 2x^2 - x + 2

One thing you can do is factor by grouping.

f(x) = x^2(x - 2) - (x - 2)

Now, group by factoring (x - 2) out of this.

f(x) = (x - 2)(x^2 - 1)

Factor the second part as a difference of squares.

f(x) = (x - 2)(x - 1)(x + 1)

Hmm... plugging in -1 and 1 should have given you 0, and it should. Let's try it out.

f(1) = 1^3 - 2(1)^2 - 1 + 2
f(1) = 1 - 2(1) - 1 + 2
f(1) = 1 - 2 - 1 + 2
f(1) = 0

Answer 2

use synthetic division
possibles are [+ or- 1,2]
1] 1 -2 -1 2
1 -1 -2
1 -1 -2 0 so, (x-1) is a zero because there is no remainder (x-1)(x^2-x-2)
try -1
-1] 1 -1 -2
-1 2
1 -2 0 so this is also one, (x+1)(x-1)(x-2)
the easiest way is to plug these into a graphing calculator and see where the line crosses the x axis and these will be your zeroes
The roots are [1,-1,2]

Answer 3

From Puggy's answer you can see that 2, 1 and -1 are zeros of the function
Here's another way to do this when the factoring isn't so simple:
Since the function is a polynomial with integer coefficients, IF the polynomial has integer roots, this roots are divisors of the independent term (the one with no x, in this case 2). This you already knew it, because it's what you tried to do.
But here's another property of polynomials with integer coefficients: IF they have rational zeros, then the numerator of the fraction is a divisor of the independent term and the denominator of the fraction is a divisor of the main coefficient (the one in the higher power of x, in this case 1 for x^3)

Answer 4

Try again, the answers are -1,1, and 2

(-1)^3 - 2(-1)^2 - (-1) + 2
-1 -2 + 1 + 2
-3 + 3
0

(1)^3 - 2(1)^2 - 1 + 2
1 - 2 - 1 + 2
-1 + 1
0

(2)^3 - 2(3)^2 - 2 + 2
8 - 8 - 2 + 2
0 + 0
0

Answer 5

f(x)= X^3-2X^2 - X +2
f(1)=1^3-2*1^2-1+2=1-2-1+2=0
so x=1 is a root. next use synthetic division

x^2 -x -2
x-1\x^3-2x^2-x+2
-(x^3-x^2)
-x^2 - x
-(-x^2+x)
-2x +2
-(-2x+2)
0
Now solve the quadratic
x^2 -x -2
(x-2)(x+1)

the 3 roots are 1, -1, & 2

Answer 6

thats all that you have to do, you did well, there is no rational solutions