Find all rational zeros of the polynomial?

Question

P(x)=x^3-4x^2-7x+10

Answer 1

Well, by inspection -2 is one root...

P(-2) = (-2)^3 - 4(-2)^2 - 7(-2) + 10
P(-2) = -8 - 16 + 14 + 10
P(-2) = -24 + 24
P(-2) = 0

Now take your equation and divide by (x - (-2)). In other words (x + 2):

............ 1 -6 5
1 2 ) 1 -4 -7 10
....... 1 2
....... ---------
........... -6 -7
........... -6 -12
........... ----------
............... 5 10
............... 5 10
............... ------
.................... 0

So the result is x^2 - 6x + 5

This can be factored to (x - 5)(x - 1)

This results in (x + 2)(x - 5)(x - 1)

And your roots are:
-2, 5, 1

(I don't know why I didn't notice 1 as a root... I usually check that first...)

Answer 2

-by inspection try x=1:
since P(1) = 0 that means x-1 is a factor.
-divide P(x) by (x-1)
(x^3 -4x^2 -7x +10)/(x-1) = x^2 -3x - 10
-you can factor x^2 -3x -10 into (x-5)(x+2)
-therefore, your zeros are 1, 5 and -2

Answer 3

Tip 1:
If the sum of the coeffictients of a polymer is zero then 1 is one of the zeros.

Tip 2:
If there are rational roots and the first coefficient of the polynom is 1, then the roots are all the divisors of the last coefficient.

So normally you will use Tip2 and try those roots (1,2,5,10 and the negatives). In this case use tip 1. do the synthetic division and solve by the quadratic formula.

x1 = 1
P(x) = (x -1)*(x^2 -3x - 10)
x2 = 5
x3 = -2
P(x) = (x - 1) * (x - 5) * (x + 2)

Answer 4

Start by listing all the possible rational zeros.
They are the factors of 10.
-1, 1, -2, 2, -5, 5, -10, 10
Now pick one until you find one that gives you P(x) = 0
Try x = 1
P(1) = 1^3 - 4(1^2) - 7(1) + 10
P(1) = 1 - 4 - 7 + 10 = 0
Since P(1) = 0 that means x-1 is a factor.
Divide P(x) by (x-1)
(x^3 -4x^2 -7x +10)/(x-1) = x^2 -3x - 10
You can factor x^2 -3x -10 into (x-5)(x+2)
Therefore, your zeros are 1, 5 and -2